Let $\sigma(x)$ be the sum of the divisors of $x$.
Here is my original question:
If $\bigg(\sigma(b^2) - b^2\bigg) \mid \bigg(b^2 - 1\bigg)$, is $$\frac{b^2 - 1}{\sigma(b^2) - b^2} + 1$$ always an odd prime for $b$ odd?
Preliminary computations using Mathematica show that this must be true. Is it?
Added March 18 2017
When $b = p^k$ where $p$ is an odd prime and $k \geq 1$, we have $$\frac{b^2 - 1}{\sigma(b^2) - b^2} + 1 = \frac{\sigma(b^2) - 1}{\sigma(b^2) - b^2} = \frac{\sigma(p^{2k}) - 1}{\sigma(p^{2k}) - p^{2k}} = \frac{p\sigma(p^{2k-1})}{\sigma(p^{2k-1})} = p,$$ validating didgogns's claim in the comments. Hence, I modify my original question slightly as follows:
MODIFIED QUESTION
If $\bigg(\sigma(b^2) - b^2\bigg) \mid \bigg(b^2 - 1\bigg)$, is $$\frac{b^2 - 1}{\sigma(b^2) - b^2} + 1$$ always an odd prime for $b$ an odd composite?