Find prime numbers $p,q$ such that $p^n+p^{n-1}+...+p+1=q^2+q+1$ Let $p,q$ are prime numbers and $n$ is a even number such that : $p^n+p^{n-1}+...+p+1=q^2+q+1$

Find $p,q$?

I think :
$p^n+p^{n-1}+...+p+1=q^2+q+1\Rightarrow p^n+p^{n-1}+...+p=q(q+1)\Rightarrow p|q(q+1)\Rightarrow p|q$ or $p|q+1$
If $p|q$ then $p=q$. I can solve this
If $p|q+1$, I don't know how to do next !?

Partial answer: Suppose $p\neq q$. $q\neq 2$ and $q$ doesn't divide $n$. Assume
$$p^{n}+\cdots +p+1=q^2+q+1$$
with $n$ an even number. Using geometric sum, we have
$$\frac{p^{n+1}-1}{p-1}=q^2+q+1$$
and therefore
$$p^{n+1}-1=(p-1)(q^2+q+1)$$
If we take modulo $q$, we obtain
$$p^{n+1}-1\equiv p-1 \mod q$$
and thus, since $p\neq q$
$$p^n\equiv 1\mod q$$
Let $m$ be the multiplicative order of $p$ in the group $\mathbb{Z_q}$, then $m$ divides $n$ and $q$ (Lagrange), hence, since $n$ and $q$ are relatively prime $m=1$. This implies
$$p\equiv 1\mod q\ .$$
At this point, we return to the original equality
$$p^{n}+\cdots +p+1=q^2+q+1$$
If we take modulo $q$ at both sides, we get
$$1+1+\cdots+1\equiv 1 \mod q$$
where we sum $1$ $n+1$ times, i.e.
$$n+1\equiv 1\mod q$$
and so $n\equiv 0\mod q$, but this is absurd since $q$ doesn't divide $n$.
