You are searching for irreducible polynomials of degree $\le3$, so the elementary irreducibility test works: a quadratic or a cubic $f(x)\in\Bbb{F}_4[x]$ is irreducible if and only if it has no zeros in $\Bbb{F}_4$. This may be the conceptually simplest way of going about this task. But I want to shed a bit more light on a few aspects of the problem. Depending on how familiar you are with finite fields and their simple Galois theory the explanation below may be not as accessible as I would like it to be. Do ask for clarifications!
It is, indeed, correct that the irreducible (monic) polynomials of degrees $1$ and $2$ in $\Bbb{F}_4[x]$ are exactly the factors of $x^{16}-x=x^{16}+x$.
In this old thread we saw that over the prime field $\Bbb{F}_2$ we have the factorization
$$
x^{16}+x=x(x+1)(x^2+x+1)(x^4+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1).\qquad(*)
$$
Furthermore, those factors are irreducible in $\Bbb{F}_2[x]$. After all, by the same general result $x^{2^4}-x$ is the product of all the irreducible polynomials in $\Bbb{F}_2[x]$ of degrees dividing $4$ (you see that the degrees of the factors are $1,2$ or $4$).
When switching to $\Bbb{F}_4$ the game is to figure out how those factors, irreducible over $\Bbb{F}_2$, factor further over $\Bbb{F}_4$.
Field theory gives the following general principle. If $p(x)$ is an irreducible polynomial of degree $m$ in $\Bbb{F}_q[x]$, then any one of its zeros, denote one of them $\alpha$, generates the field $\Bbb{F}_q(\alpha)$ that is an $m$-dimensional extension field, and $p(x)$ is its minimal polynomial. From the basic facts about finite fields we know that there is only one such field (up to isomorphism), namely $\Bbb{F}_{q^m}$.
This applies irrespective of whether $q$ is a prime or a prime power.
To apply this I first need to give a precise description of $\Bbb{F}_4$. Undoubtedly you have seen this, so I just list the elements
$$
\Bbb{F}_4=\{0,1,\beta,\beta+1\}.
$$
The arithmetic is determined by the rules $1+1=0$ and $\beta^2=\beta+1$, in line with the fact that $\beta$ is one of the zeros of the polynomial $x^2+x+1$, irreducible over $\Bbb{F}_2$. Its Galois conjugate $\beta^2$ shares that minimal polynomial, so over $\Bbb{F}_4$ that quadratic factors as
$$
x^2+x+1=(x-\beta)(x-\beta^2)=(x-\beta)(x-[\beta+1]).
$$
This was to be expected because the zeros of a quadratic, irreducible over $\Bbb{F}_2$ belong to the quadratic extension of that base field, and that's precisely $\Bbb{F}_4$.
What about the degree four factors in $(*)$? They are irreducible over $\Bbb{F}_2$, so any zero of any of them generates the degree four extension field $\Bbb{F}_{16}$. Here $16=4^2$, so $\Bbb{F}_{16}$ is a degree two extension of the field $\Bbb{F}_4$. Therefore the minimal polynomial of each and every one of those zeros has degree two over $\Bbb{F}_4$. In other words, all those quartics split into products of two irreducible quadratics from $\Bbb{F}_4[x]$.
There are several ways of finding those factors. Now that we know the factors to be quadratic we could simply use the method of unknown coefficients, and work out the resulting systems of equations. Alternatively we can use the (also known) arithmetic of $\Bbb{F}_{16}$, when we have the benefit of explicit descriptions of those zeros. The price we need to pay for that is to also describe the field $\Bbb{F}_{16}$ in more detail. I shamelessly reuse another old answer of mine to that end. So let $\gamma\in\Bbb{F}_{16}$ be a zero of $x^4+x+1$, when we can write the elements of $\Bbb{F}_{16}$ as polynomials of $\gamma$. In the linked answer I show that $\gamma$ is a primitive root of unity of order fifteen. As $\beta$ is a root of unity of order three, it follows that, as an element of $\Bbb{F}_{16}$, we must have $\beta=\gamma^5=\gamma^2+\gamma$ or $\beta=\gamma^{10}=\gamma^2+\gamma+1$.
Actually we can simply make a pick here because we don't have big superfield containing all the fields $\Bbb{F}_{2^m}$ at hand. So, again following the linked thread, I simply declare that $\beta=\gamma^2+\gamma$. The other choice differs from this by an automorphism of $\Bbb{F}_4$, so we can choose freely (but only once — from this point on the relation between $\gamma$ and $\beta$ is locked in place).
In the linked thread I already showed that over $\Bbb{F}_4$ we have the factorization
$$x^4+x+1=(x^2+x+\beta)(x^2+x+\beta^2).$$
The relation $\gamma^2+\gamma=\beta$ shows that $\gamma$ is a zero of the first factor. By Galois theory, the other zero is $\gamma^4=\gamma+1$, whereas $\gamma^2$
and $\gamma^8=\gamma^2+1$ are the zeros of the latter factor.
The quartic $x^4+x^3+1$ is the reciprocal polynomial of $x^4+x+1$, so its zeros are the conjugates of $1/\gamma=\gamma^{14}$. But if $\gamma$ is a zero of $f(x)=x^2+x+\beta$, then $1/\gamma$ is a zero of the reciprocal polynomial
$$
\tilde{f}(x):=x^2f(\frac1x)=\beta x^2+x+1.
$$
We want to use monic polynomials as minimal polynomials, so we multiply that polynomial with $1/\beta=\beta^2$, and arrive at the minimal polynomial
$$
\frac1{\beta}\tilde{f}(x)=x^2+\beta^2x+\beta^2.
$$
Galois theory tells us that the zeros of this polynomial are $1/\gamma$ and $1/\gamma^4$. As the automorphism of $\Bbb{F}_4$ simply interchanges $\beta$ and $\beta+1=\beta^2$, we see that Galois conjugate of that polynomial
$$
x^2+\beta x+\beta
$$
must be the other factor of $x^4+x^3+1=(x^2+\beta^2x+\beta^2)(x^2+\beta x+\beta)$. You can verify that factorization as an exercise to become fluent with the arithmetic of $\Bbb{F}_4$.
We still have the irreducible quartic $x^4+x^3+x^2+x+1$. It is a factor of $x^5-1$, so its roots are the fifth roots of unity in $\Bbb{F}_{16}$, that is $\gamma^3,\gamma^6,\gamma^9,\gamma^{12}$. Of these $\gamma^3$ and $\gamma^{12}$ are also Galois conjugates over $\Bbb{F}_4$. Therefore we expect $(x-\gamma^3)(x-\gamma^{12})$ to have coefficients in the subfield $\Bbb{F}_4$. Using the table from the linked thread we can expand the product as follows
$$
\begin{aligned}
(x-\gamma^3)(x-\gamma^{12})&=x^2+[\gamma^3+\gamma^{12}]x+\gamma^{15}\\
&=x^2+[\gamma^3+(\gamma^3+\gamma^2+\gamma+1)]x+1\\
&=x^2+(\gamma^2+\gamma+1)x+1\\
&=x^2+\beta^2x+1.
\end{aligned}
$$
Again, by Galois conjugation, the other factor (with zeros $\gamma^6$ and $\gamma^9$) must be $x^2+\beta x+1$, so the last quartic factors as
$$x^4+x^3+x^2+x+1=(x^2+\beta^2x+1)(x^2+\beta x+1)$$
completing the first task.
On with finding the irreducible cubics in $\Bbb{F}_4[x]$. Those will have zeros in the cubic extension $\Bbb{F}_{64}[x]$. Sixty of its elements are outside $\Bbb{F}_4$. Each irreducible cubic in $\Bbb{F}_4[x]$ has three of them as zeros, so we can already determine that there will be $60/3=20$ irreducible cubics. I hope I will be excused from not producing a comprehensive list!
To carry out the calculation as in the above case of quadratics I would need a discrete logarithm table of $\Bbb{F}_{64}$. Alas, I don't have one at hand. So, instead, I explain a few points and examples.
The easiest observation to make is that the irreducible cubics in $\Bbb{F}_2[x]$ both remain irreducible over $\Bbb{F}_4$. As $2^3=8$ those are found from the factorization
$$
x^8-x=x(x-1)(x^3+x+1)(x^3+x^2+1)\in\Bbb{F}_2[x].
$$
The reason is that the zeros of those cubics are elements of $\Bbb{F}_8$. And $\Bbb{F}_{64}$ is the smallest field containing copies of both $\Bbb{F}_4$ as well as $\Bbb{F}_8$ ($64$ is the smallest power of two that is a power of both $4$ and $8$). So adjoining any such zero $\delta$ to $\Bbb{F}_4$ necessarily gives us the field $\Bbb{F}_{64}$ and consequently the minimal polynomial of $\delta$ over $\Bbb{F}_4$ must be cubic.
That's two out of twenty :-/
Again reusing the tables in the linked thread, I will denote by $\alpha$ one of the zeros of $x^3+x+1$ in $\Bbb{F}_8$. We can think of both $\alpha$ and $\beta$ as elements of $\Bbb{F}_{64}$. We see that their sum $\xi:=\alpha+\beta$ is one of the elements in $\Bbb{F}_{64}\setminus\Bbb{F}_4$. So it has a cubic minimal polynomial $g(x)$ over $\Bbb{F}_4$. How to find it? Again, Galois theory tells us that the other zeros of $g(x)$ are $\xi^4=\alpha^4+\beta^4=\alpha^4+\beta$ and
$\xi^{16}=\alpha^{16}+\beta^4=\alpha^2+\beta$. Recall also that $\alpha,\alpha$ and
$\alpha^4$ are zeros of $x^3+x+1$. So
$$x^3+x+1=(x-\alpha)(x-\alpha^2)(x-\alpha^4)$$
tells us the values of their elementary symmetric polynomials:
$$
\begin{aligned}
e_1&:=\alpha+\alpha^2+\alpha^4&=0,\\
e_2&:=\alpha\cdot\alpha^2+\alpha\cdot\alpha^4+\alpha^2\cdot\alpha^4&=1,\\
e_3&:=\alpha\cdot\alpha^2\cdot\alpha^4&=1.
\end{aligned}
$$
By expanding
$$
\begin{aligned}
g(x)&=(x-[\alpha+\beta])(x-[\alpha^2+\beta])(x-[\alpha^4+\beta])\\
&=x^3-[\alpha+\alpha^2+\alpha^4+3\beta]x^2\\
&+[\alpha\cdot\alpha^2+\alpha\cdot\alpha^4+\alpha^2\cdot\alpha^4+2(\alpha+\alpha^2+\alpha^4)\beta+3\beta^2]x\\
&-[\alpha\cdot\alpha^2\cdot\alpha^3+(\alpha\cdot\alpha^2+\alpha\cdot\alpha^4+\alpha^2\cdot\alpha^4)\beta+(\alpha+\alpha^2+\alpha^4)\beta^2+\beta^3]\\
&=x^3+(e_1+\beta)x^2+(e_2+\beta^2)x+(e_3+e_2\beta+e_1\beta^2+\beta^3)\\
&=x^3+\beta x^2+\beta^2x+(1+\beta+\beta^3)\\
&=x^3+\beta x^2+\beta^2x+\beta,
\end{aligned}
$$
as $\beta^3=1$. The Galois conjugate of this then yet another irreducible cubic
$x^3+\beta^2x^2+\beta x+\beta^2$. Four down, sixteen to go.
The last trick I show is about factoring the cyclotomic polynomial
$$
\Phi_9(x)=(x^9-1)/(x^3-1)=x^6+x^3+1.
$$
Its roots are units of unity of order nine. As $2^6$ is the lowest power $2^m$ such that $9\mid 2^m-1$, we can conclude that $\Phi_9(x)$ is irreducible over $\Bbb{F}_2$. Hence any of its roots generates $\Bbb{F}_{64}$ and has a cubic minimal polynomial over $\Bbb{F}_4$. If $\zeta$ is one of those roots, the others are $\zeta^2$, $\zeta^4$, $\zeta^8$, $\zeta^{16}=\zeta^7$ and $\zeta^{32}=\zeta^5$.
Of these $\zeta$, $\zeta^4$ and $\zeta^{16}=\zeta^7$ are Galois conjugates over $\Bbb{F}_4$ and hence share a minimal polynomial, whereas the other roots are their reciprocals.
We see that
$$
(x-\zeta)(x-\zeta^4)(x-\zeta^7)=x^3+\cdots+\zeta^{1+4+7},
$$
where the constant term $\zeta^{1+4+7}=\zeta^{12}=\zeta^3$ is a root of unity of order three. The constant term of the other factor is $\zeta^{2+5+8}=\zeta^{-3}$.
Therefore these constant terms are each others reciprocals. We can conclude that one of them is $\beta$ and the other is $1/\beta=\beta^2$. We cannot tell which is which, because we did not specify $\zeta$ uniquely. Anyway, one of the factors of $\Phi_9(x)$ has the form
$$m(x):=x^3+c_1x^2+c_2x+\beta$$
with some unknown coefficients $c_1,c_2$ from $\Bbb{F}_4$. The other must then be the reciprocal up to a constant factor
$$
r(x):=\frac1{\beta}x^3m(\frac1x)=x^3+\frac{c_2}{\beta}x^2+\frac{c_1}{\beta}x+\frac1{\beta}.
$$
We have the factorization
$$
\Phi_9(x)=x^6+x^3+1=m(x)r(x),\qquad(**)
$$
and use that to determine $c_1$ and $c_2$. Expanding the right hand side of $(**)$ and comparing the $x^5$-terms we arrive at the equation $c_2+c_1/\beta=0$, which gives us $c_1=c_2\beta$. Plugging that in and similarly comparing the $x^3$-terms gives the equation
$$
1=\beta^2+\beta+c_2^2\beta^2+c_2^2\beta=1+c_2^2.
$$
From the we can solve $c_2=0$. At this point I feel a bit silly, because the factorization
$$
x^2+x+1=(x+\beta)(x+\beta^2)
$$
automatically yields also the factorization
$$
x^6+x^3+1=(x^3+\beta)(x^3+\beta^2).
$$
Nevertheless, that kind of thinking may help you with some other problem.
Six down, fourteen to go :-)
You can continue. And also use the fact that if $h(x)$ is an irreducible cubic, so is $h(x+z)$ for any non-zero constant $z\in\Bbb{F}_4$. As is the reciprocal $\tilde{h}(x)=x^3 h(1/x)$. Have fun!
