Let $p \in \mathbb{P}$ be a prime and suppose that an integer $e > 1$ is given such that the polynomial $s_e = 1 + x + \dots + x^{e-1}$ is irreducible in $\mathbb{F}_p[x]$.
My question is the following: what are the maximal ideals in $\mathbb{F}_p[x,y]$ that contain the principal ideal $\langle (xy)^e - 1 \rangle$?
As explained in the comments by JacobFG, the maximal ideals are gotten as kernels of the evaluation map at a fixed zero $P=(x_1,y_1)$ of $(xy)^e-1$. The Galois conjugates of $P$ give rise to the same kernel. With $x_1,y_1$ algebraic over $\Bbb{F}_p$, the Galois action is gotten by iterating the Frobenius automorphism.
Let's consider this one element $x_1\in\overline{\Bbb{F}_p}, x_1\neq0,$ at a time. Taking all the Galois conjugates together, this amounts fixing the minimal polynomial $m(x)$ over the prime field. We arrive at the following answer.
Observation. Consider an irreducible polynomial $m(x)\neq x$ of degree $n$. Let $\ell=\operatorname{lcm}(e-1,n)$, $m=\ell/n$, $k=(e-1)/m$. Then there are $1+k$ maximal ideals of $\Bbb{F}_p[x,y]$ containing $\langle m(x),(xy)^e-1\rangle$.
Proof. One of the maximal ideals comes from $y_1=1/x_1$. In this case the minimal polynomial of $y_1$ is the reciprocal $\tilde{m}(y)=y^n m(1/y)$. This maximal ideal is generated by $m(x),\tilde{m}(y)$ and $xy-1$.
When $x_1y_1=\zeta$ is a primitive root of unity of order $e$ we need to be a bit more careful. The assumption of $s_e$ being irreducible implies that $[\Bbb{F}_p(\zeta):\Bbb{F}_p]=e-1$. As $\zeta\in\Bbb{F}_p(x_1,y_1)$, it follows that $[\Bbb{F}_p(x_1,y_1):\Bbb{F}_p]=\ell$. The set of conjugate points $$S=\{(x_1^{p^i},y_1^{p^i})\mid 0\le i<\ell\}$$ has $\ell$ elements as the Frobenius automorphism gives repetitions only after the $\ell$th iteration. As $x_1$ has exactly $n$ conjugates, this means that in $S$ there are $m=\ell/n$ points sharing the first coordinate $x_1$. The matching $y$-coordinates are $y_1^{p^j}$ with $n\mid j$, and the matching roots of unity are $\zeta^{p^j},n\mid j$. This applies to any of the primitive roots of unity of order $e$. Hence the $e-1$ choices of $\zeta$ (or $y_1$) are partitioned into $(e-1)/m$ subsets of conjugate points. If we let $y_1$ range over the set of representative of those subsets, we get the maximal ideals generated by $m(x)$ and $g(x,y)\in\Bbb{F}_p[x][y]$ with the latter polynomial of degree $m$ in $y$ determined by the requirement that $g(x_1,y)$ shoud be the minimal polynomial of $y_1$ over $\Bbb{F}_p(x_1)$. We have accounted for all the solutions of $(x_1y)^e-1$, so the list is compelete.
As a small example, let's take a look at the case $p=2, e=5$, $m(x)=x^2+x+1$. The coset of $2$ is a generator of $\Bbb{Z}_5^*$, so $s_5=x^4+x^3+x^2+x+1$ is irreducible over $\Bbb{F}_2$. I will reuse the tables of the field $\Bbb{F}_{16}$ from here, so in what follows $\gamma$ satisfies the equation $\gamma^4+\gamma+1=0$. We see that $x_1=\gamma^5$ generates the subfield $\Bbb{F}_4$, and its minimal polynomial is $x^2+x+1$. If $y_1=\gamma$, then we have $x_1y_1=\gamma^6$, which is one of the primitive fifth roots of unity. Therefore $P=(x_1,y_1)$ is a solution of $(xy)^5-1$.
By elementary Galois theory, the minimal polynomial of $y_1$ over $\Bbb{F}_4=\Bbb{F}_p(x_1)$ is $$(y-y_1)(y-y_1^4)=(y+y_1)(y+y_1+1)=y^2+y+(y_1^2+y_1)=y_2+y+x_1.$$ The conclusion is that $$m_P=\langle x^2+x+1,y^2+y+x\rangle$$ is the kernel of the evaluation map at $P$. The conjugate points $$(x_1,y_1),(x_1^2,y_1^2), (x_1,y_1^4), (x_1^2,y_1^8)$$ all share this same kernel.
Of the fifth roots of unity and our choice of $x_1$ this set "covered" $\gamma^6=x_1y_1$ and $\gamma^9=x_1y_1^4$, conjugate over $\Bbb{F}_4=\Bbb{F}_2(x_1)$. To get the missing pair we need to use $y_1=\gamma^7$ instead, when $Q=(x_1,y_1)$ is again on the curve $(xy)^5=1$. The minimal polynomial of that $y_1$ is $$ (y-\gamma^7)(y-\gamma^{28})=y^2+(\gamma^2+\gamma)y+(\gamma^2+\gamma+1)= y^2+x_1y+(x_1+1).$$ So the third maximal ideal containg $x^2+x+1$ is $$ m_Q=\langle x^2+x+1,y^2+xy+x+1\rangle. $$
