Find all primes p,q and even n>2 such that $\displaystyle p^n + p^{n - 1} + ... + 1 = q^2 + q + 1$.

My idea:

$\displaystyle p(p^{n - 1} + ..... + 1) = q(q + 1)$.

Let k be the largest power of p dividing q+1, then

$\displaystyle p^{n - 1} + ..... + 1 = qcp^{k - 1}$

If k>1 then the RHS is divisible by p and LHS is not(I am using n>2)

so $\displaystyle q+1 = pc$ for some $\displaystyle c \in \mathbb{N}$

It looks like I have deduced something, but then what do we do??

Or is this approach useless??