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

Attempt
The first thing I would do is simplify the geometric series to $\dfrac{p^{n+1}-1}{p-1} = q^2+q+1$. I was thinking from here we could use a modular arithmetic argument, perhaps using FLT. It is unclear to me though how to proceed since we are working with two different primes.
 A: Subtracting $1$ from both sides and factoring yields
$$
  p(p^{n-1}+p^{n-2}+\dotsb+1)=q(q+1).
$$
Since $p$ and $q$ are primes, this leads to certain conclusions. Is that enough to go on?
A: The solutions can be found deriving a bounding condition on $p$ and $n$. Here's another approach:
Manipulate the equation into $$p\left(p^{n-1}+p^{n-2}+\cdots+1\right)=p\frac{p^n-1}{p-1}=q(q+1) \tag{$\star$}$$ to see there must exist an integer $a>1$ (since $n>2$) such that $$q+1=ap \tag{1}$$and $$ \frac{p^n-1}{p-1}=aq. \tag{2}$$ Subtract $1$ from both sides of $(2)$ to get $$\frac{p^n-p}{p-1}=p\frac{p^{n-1}-1}{p-1}=aq-1, $$ and summing this to $(1)$ we arrive at $$p\left(a+\frac{p^{n-1}-1}{p-1}\right)=q(a+1),$$ whence $a=bp-1$ for some $b\in\mathbb{N^+}$. Substituting into $(\star)$ we finally obtain $$\begin{align}\sum_{k=0}^{n-1}p^k=(bp-1)(p(bp-1)-1)&=(bp-1)(bp^2-p-1) \\ &=b^2p^3-2bp^2-(b-1)p+1\end{align}$$ and thus $$\begin{align}\sum_{k=0}^{n-2}p^k=b^2p^2-2bp-b+1 \end{align}$$ $$\sum_{k=1}^{n-2}p^k=b^2p^2-2bp-b, \tag{3}$$ which shows that $b$ is both a multiple and a divisor of $p$, i.e. $b=p$. Hence $(3)$ is equivalent to $$\begin{align}\sum_{k=0}^{n-3}p^k&=p^3-2p-1 \\p\left(p^2-2-\sum_{k=0}^{n-4}p^k\right)&=2,\end{align}$$wherefrom we conclude $(n,p,q)=(4,2,5).$
A: Suppose that $n=2k$ for some integer $k>1$.  Then we have $q(q+1)=p\left(\frac{p^{2k}-1}{p-1}\right)$.  As $k>1$, $p\neq q$, so that $q$ divides $p^{2k}-1=\left(p^k+1\right)\left(p^k-1\right)$.  That is, $q$ divides either $p^k-1$ or $p^k+1$.  In any case, $q\leq p^k+1$.  Therefore, $$p^{2k}+p^{2k-1}+2< p^{2k}+p^{2k-1}+\ldots+p=q^2+q\leq \left(p^k+1\right)^2+\left(p^k+1\right)=p^{2k}+3p^k+2\,.$$
Ergo, $p^{2k-1}<3p^k$, or $p^{k-1}<3$.  Consequently, $k=2$ and $p=2$.  Hence, the only solution $(p,q,n)$ to $p^n+p^{n-1}+\ldots+p+1=q^2+q+1$ with $p,q,n\in\mathbb{N}$ such that $p$ and $q$ are primes and that $n>2$ is even is $(p,q,n)=(2,5,4)$.
In fact, if $n$ is allowed to be an odd integer greater than $2$, then it follows that $p=2$ (otherwise $p^n+p^{n-1}+\ldots+p+1$ is even, but $q^2+q+1$ is odd).  That is, we are looking for a prime $q$ such that $q^2+q=2^{n+1}-2$, or $(2q+1)^2=2^{n+3}-7$.  As $n$ is odd, $n+3$ is even.  That is, $$\left(2^{\frac{n+3}{2}}-2q-1\right)\left(2^{\frac{n+3}{2}}+2q+1\right)=7\,.$$  Hence, $2^{\frac{n+3}{2}}-2q-1=1$ and $2^{\frac{n+3}{2}}+2q+1=7$.  Consequently, $2^{\frac{n+3}{2}}\leq 6<2^3$.  Thus, $n<3$, which is a contradiction.
In conclusion, the only solution $(p,q,n)\in\mathbb{N}^3$ to $p^n+p^{n-1}+\ldots+p+1=q^2+q+1$ with $p$ and $q$ being prime is $(p,q,n)=(2,5,4)$.
