(Note: This question is an improved version of and has been cross-posted from [this MSE post](https://math.stackexchange.com/questions/1671836).)

Let $\sigma(x)$ denote the *sum of the divisors* of $x$.  If $\sigma(x) = 2x - 1$, then we call $x$ an *almost perfect number*.

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r)$ with $r \geq 1$, then $M = {2^r}{b^2}$ is an even almost perfect number.

Now, if $\sigma(2^r)$ is prime, then ${2^r}{\sigma(2^r)}$ is an *even perfect number*.

In an answer to the [linked MSE question](https://math.stackexchange.com/questions/1671836), [Giovanni Resta](https://math.stackexchange.com/users/312312) asserts that:
> If $b = 3^k$ for $k > 1$ then $q = 2$.  Up to ${10}^8$ there are no other values of $b$ that make $q$ prime.

However, by work of [Antalan](http://dx.doi.org/10.1063/1.4882588), it is known that $3 \nmid b$.

Hence, I cannot help but ask:

>> **(1)** Is it possible to prove that if $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is prime, then necessarily $b = 3^k$ with $k > 1$?  If it is not possible by current mathematical methods, what is the main obstruction?

>> **(2)** If $b$ is an odd composite and $r \geq 1$, is it possible to prove that $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} \neq \sigma(2^r)$?  If it is not possible by current mathematical methods, what is the main obstruction?