What are probable candidates for the numbers whose $\sigma(n)$ (sum of divisors) is prime?
I know that the list of probable candidates include perfect squares and odd powers of 2 (specifically only 2). Source: https://oeis.org/A023194
But how do we go about in proving this?
The function $\sigma$ is multiplicative: $\sigma(nm) = \sigma(n) \sigma(m)$ if $n, m$ are relatively prime. Clearly $\sigma(n) > 1$ if $n > 1$. It follows that any $n$ with $\sigma(n)$ prime must be a prime power $\sigma(n) = p^k$. For such an $n$, we have $\sigma(n) = p^k + \cdots + p + 1$.
If $p$ is odd, then $\sigma(n) \equiv k + 1\pmod{2}$, forcing $k$ to be even (as clearly $\sigma(n) \not = 2$).
If $p$ is even, then $\sigma(n) = 2^{n + 1} - 1$. Since $(2^m - 1) | (2^{n+1} - 1)$ if $m | (n + 1)$, it follows that $n + 1$ must be prime. In particular, $n = 2^k$ with $k = 1$ or $k$ even (not odd: the first few values for which $\sigma(2^k)$ is prime are $2^1, 2^2, 2^4, 2^6, 2^{12}, \dots$).