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 $q = 2^{r + 1} - 1$?

Question

(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?

Answer 1

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.

Answer 2

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 $3 \nmid b$.