Let $F_p$ be the prime field of order $p$, $\overline{F_p}$ be its algebraic closure, and $n$ be an integer such that $\gcd(n,p)=1$. Consider the following three polynomials in $F_p[x,y]$: $$ p_1(x,y)=x^n - 1~, \\ p_2(x,y)=y^n - 1~, \\ p_3(x,y)=x(y-1)^2+y(x-1)^2~. $$ It is clear that $(1,1)$ is a common zero of these polynomials, so I call this the trivial solution. Say $(x_0,y_0)\in\overline{F_p}^2$ is a common zero of the above system of polynomials over $\overline{F_p}$. We can generate seven other solutions using the following transformations: $$ x_0 \leftrightarrow x_0^{-1}~,\qquad y_0 \leftrightarrow y_0^{-1}~,\qquad x_0 \leftrightarrow y_0~, $$ where the first two transformations are well-defined because $n$th roots of unity are nonzero. Note that $(1,1)$ is the only solution that's invariant under these transformations.
I want to show that
When $p>2$, there are infinitely many integers $n$ such that $(1,1)$ is the only common zero of these polynomials over $\overline{F_p}$.
I excluded the case $p=2$ because we can always choose $x_0 = y_0$ to be any $n$th root of unity, and it would satisfy $p_3(x,y)=0$.
One way to show this would be to compute the Groebner basis of the ideal generated by the above three polynomials. In lexicographic ordering, if the Groebner basis is $(x-1,y-1)$, then $(1,1)$ is the only solution. I computed it for some small $p$ and small prime $n\ne p$ (I give some examples below), and it seems like whenever $n$ is not of the form $\frac{p^k+1}{2}$ for some integer $k$, then the Groebner basis is $(x-1,y-1)$. I don't know how to show this for all $p$, and all prime $n$ not of the form $\frac{p^k+1}{2}$.
Since the above system is zero-dimensional (because $p_1$ and $p_2$ have finite number of solutions), I thought I could instead try the characteristic set method of Wu or Ritt. I computed the characteristic sets for some small $p$ and $n$ but couldn't see any pattern that could be generalised.
Here are some examples where I restrict $n\ne p$ also to be a prime less than $100$. When $p=3$, I could show by either method that $(1,1)$ is the only common zero whenever $n\ne 41 = \frac{3^4+1}{2}$. [Note that $\frac{3^3+1}{2}=14$ is not a prime, and $\frac{3^2+1}{2}=5$ and $\frac{3^1+1}{2}=2$ are small enough that the Groebner basis is still $(x-1,y-1)$.] Similarly, when $p=5$, $n\ne 13 = \frac{5^2+1}{2}$, and when $p=11$, $n\ne 61 = \frac{11^2+1}{2}$.
I would really appreciate any approaches that would prove/disprove the above statement for any $p$, or even for some fixed small $p$, say $p=3$.