If $b$ is an odd composite number and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$? (Note: An improved version of this question has been cross-posted to MO.)
Let $\sigma(X)$ be the sum of the divisors of $X$.  For example, $\sigma(2) = 1 + 2 = 3$, and $\sigma(4) = 1 + 2 + 4 = 7$.
My question is:

If $b$ is an odd composite and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, what happens when $q = 2^{r + 1} - 1$ (with $r \geq 1$)?

Without the restriction on $q = 2^{r + 1} - 1$ being prime, I only know that $M = {2^r}{b^2}$ must be an even almost perfect number.  (That is, it must satisfy $\sigma(M) = 2M - 1$.)
I guess my question can be rephrased as follows:


(1) If $b$ is an odd composite, how often is $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ a prime number?
(2) If $b$ is an odd composite, and $\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = q$ is a prime number, how often is $q$ a Mersenne prime?



The only thing that I can say is 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.
