Let $p^k m^2$ be an odd perfect number with special prime $p$.
This answer says something about the Descartes-Frenicle-Sorli Conjecture on odd perfect numbers (i.e. the prediction $k=1$).
Assume that $p^k < 2m - 1$. We will show that $k \neq 1$.
Then $$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{2pm - p - 1}{2(p - 1)}$$ which is less than $m$ if $p > 2m - 1$. By assumption, we have $$p \leq p^k < 2m - 1$$ which means that $$m < \frac{\sigma(p^k)}{2}.$$ We thus obtain $$p^k < 2m - 1 < \sigma(p^k) - 1 < \sigma(p^k)$$ which implies that $k \neq 1$ holds, under the assumption $p^k < 2m - 1$.
Assume that $p^k > 2m - 1$. We will show that $k \neq 1$.
Then $$\frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} > \frac{2pm - p - 1}{2(p - 1)}$$ which is more than $m$ if $p < 2m - 1$.
Note that $$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} > m$$ does hold, under the assumption $p^k > 2m - 1$.
By assumption, we have $$2m - 1 < p^k$$ which means that $$p < 2m - 1 < p^k,$$ or $k \neq 1$ holds, under the assumption $p^k > 2m - 1$.