Decompose $n: \ \prod_i p_i^{e_i}$; then $\sigma(n) = \prod_i\left(\sum_{j=0}^{e_i} p_i^{j}\right)$ Immediately, we see that $p_i\not=2$, hence each $p_i$ is odd $(1)$. Also, it is clear that \[\sigma(n) = 2^k \implies \sum_{j=0}^{e_i} p_i^{j}=2^{k_i}\] Now $e_i+1$ is a power of two; indeedsuppose for contradiction that it wasn't. Then pick an odd prime $q|e_i+1$: \[2^{k_i}=\sum_{j=0}^{e_i} p_i^{j}=\left(\sum_{j=0}^{q-1} p_i^{j}\right)\left(\sum_{j=0}^{\frac{e_i+1}{q}-1} p_i^{qj}\right)\] However, this yields a contradiction, since the quantity $\left(\sum_{j=0}^{q-1} p_i^{j}\right)$ is the sum of an odd number of odd terms (by $(1)$), and hence clearly odd. Lastly, it is clear to see from above (and well-known) that the \[\text{number of divisors } = \prod_i (e_i+1)\] which must be a power of two, since each $e_i+1$ is a power of two.