You can use some good reciprocal polynomial arithmetic to do this problem from first principles. See here and here for a more general treatment of this problem. Note that nothing more than some basic polynomial coefficient matching and the concept of a reciprocal polynomial is required here.
Given a polynomial $f$, the "reciprocal" polynomial $\bar{f}$ is defined as the polynomial obtained by reversing the coefficients. More precisely, $\bar{f} = x^{\deg f} f(\frac 1x)$. For example, the reciprocal of $2x^3+3x-1$ is $-x^3+3x^2+2$. An important observation resulting from the above expression is that $\overline {fg} = \bar{f}\bar{g}$ for all $f,g$ with constant term non-zero.
An interesting observation is that if $f$ has non-zero constant term (i.e. $0$ is not a root of $f$) then the roots of $\bar{f}$ are exactly the reciprocal of the roots of $f$ (with multiplicity). This is obvious from the expression for $\bar{f}$.
Let $f(x) = x^7+x^4+1$. First, write $x \to -x$ and multiply by $-1$ to get $g(x) = x^7-x^4-1$. We then have $\bar{g}(x) = -x^7-x^3+1$. Note that $g$ and $\bar{g}$ don't have any common roots. To see why, let $\alpha$ be a common root. Then, $\alpha^7 = 1-\alpha^3$ from $\bar{g} = 0$, and $1+x^4$ from $g=0$. Therefore, $1-\alpha^3= 1+\alpha^4$ which forces $-\alpha^3 =\alpha^4$ which forces $\alpha = 0, - 1$, neither of which are roots.
Suppose that $g = pq$ for non-constant polynomials $p,q$. Then, consider $\bar{g} = \bar{p}\bar{q}$. Note that $g\bar{g} = pq\bar{p}\bar{q} = p\bar{q} \bar{p}q = (p\bar{q}) \overline{(p\bar{q})}$. Let's look at the polynomial $h = p\bar{q}$. Clearly $h \neq \pm g$ : If so, then $q = \pm \bar{q}$ and therefore is a factor of both $g$ and of $\bar{g}$ which cannot happen by co-primality (since $q$ is non-constant). Similarly, $h \neq \pm \bar{g}$. All in all, it is clear that $h$ is a solution to $g\bar{g} = h \bar{h}$ which isn't any of $g$ or $\bar{g}$ up to a sign.
Which is where our investigation begins : can we prove, using polynomial comparison techniques, that $g \bar{g} = h \bar{h}$ admits no solution apart from $\pm g , \pm \bar{g}$? If so, then we're done because a consequence of reducibility was the existence of such a $h$, but we're also going to contradict it as we write.
First, what's easy to compute is $g \bar{g}$ : $$
g \bar{g} = -x^{14}+x^{11}-x^{10}+3x^7-x^4+x^3-1
$$
Now assume that $h \bar{h}$ equals the above polynomial. What inferences can we make? For starters : up to a sign, clearly the degree of $h$ equals $7$, and the coefficient of $x^{14}$ is the product of the constant term and the leading term, which equals $-1$ here. For integer-coefficient polynomials, that's only possible if , up to a sign, $h$ is monic and has constant term $-1$.
So write $$
h = x^7+a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x-1 \\
\bar{h} = -x^7+a_1x^6+a_2x^5+a_3x^4+a_4x^3+a_5x^2+a_6x+1
$$
Now comes the big observation : what's the coefficient of $x^7$ in $g \bar{g}$? It's equal to $3$. Therefore, the same is expected of $h \bar{h}$. However, observe something very interesting about the coefficient of $x^7$ in $h \bar{h}$ : clearly, to get $x^7$ you combine the coefficients of $x^n$ and $x^{7-n}$ in the polynomials $h$ and $\bar{h}$. However, those coefficients equal each other!
In other words, the coefficient of $x^7$ in $h \bar{h}$ equals $$
1+a_6^2+a_5^2+a_4^2+a_3^2+a_2^2+a_1^2+1 = 3
$$
and therefore $$
a_6^2+a_5^2+a_4^2+a_3^2+a_2^2+a_1^2 = 1
$$
which can only happen if exactly one of the $a_i$ is non-zero and equals $\pm 1$, and the others all equal $0$. This observation reduces most of the coefficients to dust. It's now time to finish the problem using the first few coefficients.
Comparing $x$-coefficients of $h \bar{h}$ and $g$ (both equal $0$) gives $a_1 = a_6$. Because at most one of $a_1$ or $a_6$ can be non-zero, it follows that both must be equal to $0$. (otherwise both are non-zero, contradicting the last paragraph).
Now compare the $x^2$ coefficients. Noting that $a_1=a_6=0$, this gives $a_2-a_5 = 0$. Therefore, $a_2 = a_5$. Again, the previous paragraph's argument implies that both equal $0$.
Now compare the $x^3$ coefficients. This gives $a_3-a_4 = 1$. So either $a_3=0,a_4 = -1$ or $a_3 = 1, a_4 = 0$.
The first gives $x^7-x^4-1 = g$. The second, gives $x^7+x^3-1 = -\bar{g}(x)$. Consequently, we've shown that $h$ is either $g$ or $\bar{g}$ up to a sign, a contradiction.
Thus, $h$ is , in fact, an irreducible polynomial.
Result : Let $1<m<n, m \neq \frac n2$ and let $\delta,\epsilon$ be any of $\pm 1$. Then, $x^n+\delta x^m+\epsilon$ is irreducible if and only if it shares no root with $\epsilon x^n+\delta x^{n-m}+1$.
Proof : Pretty much what you saw here, refined. If they do share a root, well, they also share the reciprocal of that root, and therefore both have a quadratic factor $(x-l)(lx-1)$ in common for some $l$. (or linear factor, if that root is $\pm 1$).
This criterion seems to be useful for "sparse" coefficient integer polynomials, where you have very little chance of $h \bar{h} = g \bar{g}$ having another root because of coefficient constraints. As long as you can do something with the middle coefficient (which did the trick here), it seems that the argument can be extended to more complicated polynomials than mentioned in the sheet.
Also note that there is a good reason for developing this theory : Osada's paper on the Galois groups of $x^m+ax^l+b$ (see here) requires the irreducibility of this polynomial to be a necessary condition for the Galois group to be equal to $S_n$. We have just developed a necessary criterion for this to occur, and therefore are able to find Galois groups for a slightly larger and non-trivial
class of polynomials.
