Let $\sigma_{1}$ be the classical sum-of-divisors function. For example, $\sigma(12)=1+2+3+4+6+12=28$.
Here is my question:
When does $\sigma_{1}(p^{2k}) = q$ have a solution, where $p$ and $q$ are primes, and $k \geq 1$?
MY ATTEMPT
By a well-known formula for the sum of divisors of a prime power, we obtain $$\sigma_{1}(p^{2k}) = \frac{p^{2k+1} - 1}{p - 1}.$$
If the equation $\sigma_{1}(p^{2k}) = q$ holds, then the integer $$\frac{p^{2k+1} - 1}{p - 1}$$ is prime.
Since $2k+1$ is odd and $2k+1 \geq 3$, then $$\sigma_{1}(p^{2k}) = \frac{p^{2k+1} - 1}{p - 1} \geq p^2 + p + 1.$$
Equality occurs when $k=1$.
Suppose $k=1$. Then we are considering the equation $$p^2+p+1=\sigma(p^2)=q.$$
If $q=2$, then $p(p+1)=1$. A contradiction. Therefore, $q \geq 3$.
If $p=2$, then $q=7$. No contradiction. So one possibility is $(k,p,q)=(1,2,7)$.
If $p \equiv 1 \pmod 4$, then $q=p^2+p+1 \equiv 3 \pmod 4$. No contradiction.
If $p \equiv 3 \pmod 4$, then $q=p^2+p+1 \equiv 1 \pmod 4$. No contradiction.
If $p=3$, then $q=p^2+p+1=13$. No contradiction. So another possibility is $(k,p,q)=(1,3,13)$.
If $p \equiv 1 \pmod 3$, then $q=p^2+p+1 \equiv 0 \pmod 3$, which implies that $q=3$. Therefore, $p(p+1)=2$. This contradicts $p \geq 2$. Therefore, $p \not\equiv 1 \pmod 3$.
If $p \equiv 2 \pmod 3$, then $q=p^2+p+1 \equiv 1 \pmod 3$. No contradiction.
Alas, this is as far as I could go, with my elementary congruence methods. And I have only been able to consider the case $k=1$!
Added September 24 2016 Note that we necessarily have $p \neq q$, because otherwise $p^2 + 1 = 0$, which is a contradiction. Suppose further that $q < p$. Then we have $p^2 + p + 1 = q < p$, which implies $p^2 + 1 < 0$, again a contradiction. Therefore, we must have $p < q$. This holds even for $k \neq 1$.
(I put the tag diophantine-equations because I could not find an exponential-equations tag. Please feel free to remove this tag if it is not suitable for this post.)
If $2k+1 = ab$, then $p^{2k+1}-1 = (p^a)^b - 1$ is divisible by $p^a-1$ and $p^b-1$. Thus a necessary condition is that $2k+1$ is prime.
The case $p=2$ gives you the Mersenne primes.
See e.g. OEIS sequence A028491 ($n$ such that $(3^n-1)/2$ is prime) and the links and references there.