Since $\sigma(n)$ is a multiplicative function, if $\sigma(n)$ is a power of $2$ whose exponent is a positive integer, then $\sigma(p^{k-1})$ also is a power of $2$ whose exponent is a positive integer, where $p$ is a prime and $k$ an integer larger than $1$, such that $p^{k-1}\parallel n$, since $\sigma(p^{k-1})\mid\sigma(n)$ and $\sigma(n)=2^0\Leftrightarrow n=1$. Note that $\sigma(p^{k-1})=2^m$, where $m$ is a positive integer, is equivalent to $\frac{p^k-1}{p-1}=2^m$, which is equivalent to $\prod_{1<d\mid k}\Phi_d(p)=2^m$. Consider the following three cases: Case 1: $k$ has at least an odd prime divisor. Let $q$ be that odd prime divisor. Then $\Phi_q(p)=\sum_{i=0}^{q-1}p^i$ is larger than $1$ and is a power of $2$, therefore $2\mid\Phi_q(p)$. But we have $\Phi_q(p)\equiv_2\sum_{i=0}^{q-1}p^i\equiv_21+\sum_{i=1}^{q-1}p\equiv_2p(q-1)+1\equiv_21$, a contradiction. Case 2: $k=2^q$, where $q$ is an integer larger than $1$. Since $2\mid2^q$ and $4\mid2^q$, then $\Phi_2(p)=p+1$ and $\Phi_4(p)=p^2+1$ are larger than $1$ and are powers of $2$. Since $p^2+1>p+1$, then $p+1\mid p^2+1$. Since $p+1\mid p^2-1$ and $(p^2+1)-(p^2-1)=2$, then $p+1\mid2$, therefore $p+1=2$, which means $p=1$, which is not a prime, a contradiction. Case 3: $k=2$. In this case $p^{k-1}$ is a prime number and we have $p+1=2^m$, which means $p$ is a Mersenne prime. We can conclude if $\sigma(n)$ is a power of base $2$ whose exponent is a positive integer, then $n$ is the product of distinct Mersenne primes.

Clearly the first implication is trivial (because $\sigma$ is multiplicative and the sum of the divisors of a Mersenne prime is a power of two), so suppose that $\sigma(n)=2^b$, and WLOG let $n=p^{a-1}$ be a power of a prime (just factor $n$ and observe that the only divisors of a power of two are powers of two). Clearly $p\neq 2$ (or else $\sigma(n)=1+p+\cdots+p^{a-1}\equiv 1 \bmod{2}$), so suppose that $p\geq 3$, and $a\geq 2$. Case 1:$p\equiv 1 \bmod{4}$. Suppose that $2^r\|p-1$ and $2\|p+1$, so that $2^{r+1-1}\|\frac{p^2-1}{2}$. Using the Lifting lemma we conclude that $2^{r+v_2(a)}\|p^a-1$. But $p^a-1=2^b(p-1)$, and this implies that $2^{b+r}\|p^a-1$. We conclude that $r+v_2(a)=r+b\Rightarrow b=v_2(a)$, and $2^b=1+p+\cdots+p^{a-1}>p^{a-1}\geq p^{2^b-1}>2^{2^b-1}$. But this is a contradiction, as desired (because $2^b-1\geq b$ for all $b\geq 1$). Case 2:$p\equiv 3 \bmod{4}$ Suppose that $2^r\|p+1$ and $2\|p-1$, so that $2^{r+1-1}\|\frac{p^2-1}{2}$. Using the Lifting lemma we conclude that $2^{r+v_2(a)}\|p^a-1$. But $p^a-1=2^b(p-1)$, and this implies that $2^{b+1}\|p^a-1$. We conclude that $r+v_2(a)=b+1\Rightarrow 2^b=\frac{1}{2}2^r\times 2^{v_2(a)}\leq 2^{r-1}a$. Now suppose that $2^r\neq p+1$, so that $2^r\leq \frac{p+1}{2}$, and $p^{a-1}<1+p+\cdots+p^{a-1}=2^b\leq \frac{1}{4}(p+1)a$. But using Bernoulli's inequality we get that $p^{a-1}\geq 1+(p-1)(a-1)>(p-1)(a-1)$, and this is a contradiction, because $\frac{p+1}{2}\leq p-1$ for $p\geq 3$ and $\frac{a}{2}\leq a-1$ for $a\geq 2$, so that $\frac{1}{4}(p+1)a=\frac{p+1}{2}\times \frac{a}{2}\leq (p-1)(a-1)$. We conclude that $2^r=p+1\Leftrightarrow p=2^r-1$, so that $p$ is a Mersenne prime, and this implies the problem statement.