Can $\sigma(p^k)=2^n$ for some $k>1$? If $k=1$ then the only solutions to
$$
\sigma(p^k)=2^n
$$
are when $p$ is a Mersenne prime (of course $p$ is restricted to the primes). Is there a solution for larger $k$? It doesn't seem so but I can't find a proof. It's easy enough to exclude some congruence classes but that's a long way from showing that there are no examples.
 A: Clearly $p=2$ is out of the reckoning.
I use the
Fact. If $\ell+1\mid k+1$, then $\sigma(p^{\ell})\mid \sigma(p^k)$. 
This follows by cancelling a factor $p-1$ from the relation $p^{\ell+1}-1\mid p^{k+1}-1$.
Let $q$ be a prime factor of $k+1$. Because
$\sigma(p^{q-1})\mid\sigma(p^k)$
we can deduce that $\sigma(p^{q-1})$ is also a power of two. But $p$ is odd, so $$p^i\equiv1\pmod2$$ for all $i$. Hence $$\sigma(p^{q-1})\equiv q\pmod2.$$ Therefore $q=2$, and we can conclude that $k+1$ must be a power of two as it has no odd prime factors.
But, if $k+1$ is a multiple of four, we get the divisibility relation
$
\sigma(p^3)\mid \sigma(p^k)
$
implying that
$$
\sigma(p^3)=1+p+p^2+p^3=(1+p)(1+p^2)
$$
must also be a power of two. Consequently both $1+p$ and $1+p^2$ are powers of two as well.
But $1+p^2\equiv2\pmod4$ and $p>2$, so this is impossible.
A: First $p = 2$ does not work, so we can assume $p \ge 2$.
I will use Zsigmondy Theorem :

Zsigmondy theorem : if $1 \le b < a$ are coprime and $n \ge 2$, then there exist a prime $p$ such that $p | a^n - b^n$ but $p \not \mid a^k - b^k$ for $1 \le k < n$...
... except for $2^6 - 1^6$, or if $n=2$ and $a+b$ is a power of two.

Such a $p$ is called a primitive prime factor.
Before using Zsigmondy theorem, we note that for $p$ prime, for $k \in \mathbb{N}$, $\sigma (p^k) = \sum \limits_{i=0}^k p^i = \frac{p^{k+1}-1}{p-1}$.
Case 1 : $k=1$
Then, as you pointed out in your post, the only solutions are Mersenne primes (note that it is one of the exceptions to Zsigmondy theorem)
Case 2 : $k \ge 2$
Given $p>2$ prime, as we have $k+1 \ge 3$, we can apply Zsigmondy theorem : there exists a prime $q$ such that $q \mid p^{k+1}-1$ but $q \not \mid p-1$. As $p-1$ is even, $q \neq 2$, and we also have $q \mid \frac{p^{k+1}-1}{p-1}$ i.e. $q \mid \sigma(p^k)$. Thus $\sigma(p^k)$ is not a power of $2$.
Conclusion
The only solutions are Mersenne primes.

There is a way to get rid of Zsigmondy theorem : see Jyrki Lahtonen's answer.
