Your attempt is not quite right: if $f\in F[x,y]$ factors as $f=f_1f_2$, then $V(f)$ is singular along $V(f_1)\cap V(f_2)\subset\Bbb A^2$, but these subvarieties need not meet in $\Bbb A^2$ (e.g. $x(x+1)$). The fix is to consider $f^h\in F[x,y,z]$, the homogenization of $f$ with respect to $z$, and look at $V(f^h)\subset\Bbb P^2$: if $f$ factors nontrivially as $f=f_1f_2$, then $f^h=f_1^hf_2^h$, and $V(f_1^h)\cap V(f_2^h)\subset\Bbb P^2$ must be nonempty by Bezout, so if $f$ is reducible then $V(f^h)$ must be singular. We can check this by the Jacobian criterion.
Homogenizing our $f$, we get $f^h=z^{2k}+x^kz^k+y^kz^k+x^{2k}+y^{2k}$ which has Jacobian $$\begin{pmatrix} kx^{k-1}(2x^k+z^k) & ky^{k-1}(2y^k+z^k) & kz^{k-1}(x^k+y^k+2z^k) \end{pmatrix}$$ and $v\in V(f^h)$ is singular iff all the entries of the Jacobian vanish at $v$. Now we do some casework:
- If $p\mid k$, then all entries of our Jacobian vanish identically, and indeed our original polynomial is $(1+x^{k/p}+y^{k/p}+x^{2k/p}+y^{2k/p})^p$, which isn't irreducible. From now on, assume $(k,p)=1$.
- If $p=2$, then the entries of our Jacobian vanish iff $x^{k-1}z^k$, $y^{k-1}z^k$, and $x^kz^{k-1}+y^kz^{k-1}$ do. So either $x=y=0$ or $z=0$. In the first case, $[0:0:1]$ is not on our curve, but in the second case we do have singular points: $[\zeta:1:0]$ for $\zeta$ a $2k^{th}$ root of unity. From now on, assume $p\neq 2$.
- If any two of $x,y,z$ are zero, then the third must be zero as well, which gives that our curve is nonsingular. Therefore we may assume that either one or none of $x,y,z$ are zero.
- If $z=0$ and $x,y\neq 0$, then the first two entries of our Jacobian don't vanish anywhere.
- If $x=0$ and $y,z\neq 0$ then we must have $2y^k+z^k=0$ and $y^k+2z^k=0$, which implies $(2y^k+z^k)-(y^k+2z^k)=y^k-z^k=0$. For such a point to be on our curve, it must also satisfy $z^{2k}+y^kz^k+y^{2k}=0$, which gives us that $3y^{2k}=3z^{2k}=0$. When $p\neq 3$, we get that $y=z=0$ and we have don't have singular points. When $p=3$, the singular points with $x=0$ are of the form $[0:\zeta:1]$ for $\zeta$ a $k^{th}$ root of unity. The case $y=0$ and $x,z\neq 0$ follows by symmetry.
- If none of $x,y,z$ are zero, then we must have $2x^k+z^k=2y^k+z^k=x^k+y^k+2z^k=0$. But then two times the last equation minus the first two equations implies $2z^k=0$, so there are no singular points here.
What we've shown so far is that if $p\neq 2,3$ and $(k,p)=1$, then $f$ is irreducible, while if $p\mid k$ then $f$ is a $p^{th}$ power and reducible. The cases $p=2$ and $p=3$ require more investigation - our curve has singularities there, but this does not automatically imply that $f$ is reducible (consider $y^2z=x^3+x^2z$, for instance).
When $p=2$, let $\alpha\in\overline{\Bbb F_2}$ solve $t^2+t+1$. Then we have $$(\alpha x^k+\alpha y^k+1)((\alpha+1)x^k+(\alpha+1)y^k+1)=x^{2k}+y^{2k}+x^k+y^k+1,$$ and $f$ is reducible.
When $p=3$, if $k$ is even, we may find the following factorization: $$(x^k+\sqrt{2}x^{k/2}y^{k/2}+y^k-1)(x^k-\sqrt{2}x^{k/2}y^{k/2}+y^k-1)=x^{2k}+y^{2k}+x^k+y^k+1.$$
I claim that when $k$ is odd, $f$ is irreducible. By considering $f$ over $F[x^{1/2},y^{1/2}]$, we have $f$ factors as before, so it suffices to prove that $x^{2l}\pm\sqrt{2}x^ly^l+y^{2l}-1$ is irreducible in $k[x,y]$. The Jacobian of the homogenization of these polynomials are $$\begin{pmatrix} lx^{l-1}(2x^l\pm\sqrt{2}y^l) & ly^{l-1}(2y^l\pm\sqrt{2}x^l) & -2lz^{2l-1} \end{pmatrix},$$ whose entries simultaneously vanish iff $x=y=z=0$ or $2x^l\pm\sqrt{2}y^l=2y^l\pm\sqrt{2}x^l=z=0$. The first case can't happen, while the second case gives $z=0$ and $(2\pm\sqrt{2})(x^l+y^l)=0\Rightarrow x^l+y^l=0$, which is not satisfied at any point on $V(x^{2l}\pm\sqrt{2}x^ly^l+y^{2l}-z^{2l})$.
In summary: $f$ is absolutely irreducible iff $(k,p)=1$ and either 1) $p=3$ while $k$ is odd or 2) $p>3$.
