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?


 A: 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.
A: This is only a partial answer to my initial question.
Let $I(x) = \sigma(x)/x$ be the abundancy index of $x$.
If $b$ is an odd composite and
$$\dfrac{b^2 - 1}{\sigma(b^2) - b^2} = \sigma(2^r) = 2^{r+1} - 1,$$
then I know that
$$b^2 - 1 = \sigma(2^r)\sigma(b^2) - {b^2}\left(2^{r+1} - 1\right)$$
so that
$$\sigma\left({2^r}{b^2}\right) = 2^{r+1}{b^2} - 1$$
whence $M = {2^r}{b^2}$ must be an even almost perfect number that is not a power of two.
If, in addition,
$$\sigma(2^r) = 2^{r+1} - 1 = q$$
is prime, then notice that we have
$$q + 1 = \sigma(q) = 2^{r+1}$$
so that we obtain
$$I(q) = \dfrac{\sigma(q)}{q} = \dfrac{2^{r+1}}{\sigma(2^r)} = \dfrac{2}{I(2^r)},$$
whence we get $I(2^r)I(q) = I({2^r}q) = 2$ (since $\gcd(2^r, q) = \gcd(2^r, \sigma(2^r)) = 1$).  Consequently, this means that $N = {2^r}{q}$ is an even perfect number (since $r \geq 1$).  By the Euclid-Euler characterization for even perfect numbers, $q = 2^{r+1} - 1$ must be a Mersenne prime, so that $r + 1$ must be prime.
Lastly, as an aside to Giovanni Resta's earlier answer, it is known (by work of Antalan) that $3 \nmid b$.
