Are $121$ and $400$ the only perfect squares of the form $\sum\limits_{k=0}^{n}p^k$?

I've been looking for perfect squares that can be represented as $\sum\limits_{k=0}^{n}p^k$.

Of course, both $n$ and $p$ should be natural numbers larger than $1$.

Searching up to $n=100$ and $p=200$, I found only $2$ cases:

• $121=11^2=3^0+3^1+3^2+3^3+3^4=\sum\limits_{k=0}^{4}3^k$
• $400=20^2=7^0+7^1+7^2+7^3=\sum\limits_{k=0}^{3}7^k$

Is there any way to prove that there are no other cases?

Turning a comment into an answer (of sorts)....

Various papers (e.g., this one by Yann Bugeaud and Preda Mihailescu) cite papers by Nagell and Ljunggren as proving there are no perfect squares of the given form other than the ones the OP found.

As for the more general problem of representing an arbitrary perfect power, a recent paper by Michael Bennett and Aaron Levin offers this assessment in its abstract (boldface emphasis added):

The Diophantine equation ${x^n−1\over x-1} = y^q$ has four known solutions in integers $x$, $y$, $q$ and $n$ with $|x|, |y|, q \gt 1$ and $n \gt 2$. Whilst we expect that there are, in fact, no more solutions, such a result is well beyond current technology.

The other two solutions referred to are the perfect cube $7^3$ expressed as $1+18+18^2$ and $1+(-19)+(-19)^2$.

See also the MathOverflow answer by quid.