As far as I have controlled:
$\sigma(a)=2^n$, for some $n\in\mathbb N \iff $ $a$ is a product of different Mersenne primes.
The $\Leftarrow$-part is an immediate consequence of that $\sigma$ is multiplicative, but the other part seems more complicated, if it is true.
Maybe this is interesting in the view of that it is unknown if there are only a finite number of Mersenne primes or not.
Here is a solution for $n$ squarefree:
Assume the lhs is true. Then you can split $a$ up into its primes. By definition you have $\sigma(p)=p+1$ for each, thus $$\sigma(\prod_{i=1}^r p_i)=\prod_{i=1}^r (p_i+1)=2^n$$ Since $2$ is the only prime factor of the right product, every factor is at least of the "Mersenne form" $2^m-1$. If $m$ is a prime, we're done. If $m=pq$ ($p$ prime, $q$ integer) is computed then consider $$2^m-1=(2^p-1)(2^{(q-1)p}+2^{(q-2)p} +...+2^{p+1})$$ Repeat the procedure with $2^{n/p}$.
The generalization could come from the fact that $\sigma(p_n\#)=2^n$ where $p_n\#$ is the primorial, which only consists of square free primes.