This question is an offshoot of this earlier one and this other question as well.
Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m)=1$. Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
Dris proved that $p^k < m^2$ and conjectured that $p^k < m$. The first inequality, together with Ochem and Rao's lower bound for the magnitude of an odd perfect number that $p^k m^2 = n > {10}^{1500}$, implies that $m > {10}^{375}$.
Now, following the discussion in the hyperlinked questions, we have the (Diophantine) equation $$m^2 - p^k = 4z.$$
We obtain $$m^2 - 1 = p^k + (4z - 1).$$
The last equation is equivalent to $$(m+1)(m-1) = p^k + (4z - 1)$$ which implies that $${10}^{375} - 1 < m - 1 = \frac{p^k + (4z - 1)}{m + 1}$$ from which we obtain $$({10}^{375} - 1)(m + 1) < p^k + (4z - 1).$$
The last inequality implies that $$m < ({10}^{375} - 1)m < p^k + [(4z - 1) - ({10}^{375} - 1)] < p^k$$ provided that $$m^2 - p^k = 4z < {10}^{375}< m.$$ But the inequality $$m^2 - p^k < m$$ together with the inequality $$m < p^k$$ will imply that $$\frac{m^2}{2} < p^k,$$ contradicting Dris and Luca's lower bound of $\sigma(m^2)/p^k > 5$.
Added in response to a comment from MSE user mathlove
Since $$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(n)=2n=2 p^k m^2,$$ $\sigma(m^2)/p^k > 5$ implies that $\sigma(p^k)/m^2 < 2/5$, from which it follows that $$p^k < \sigma(p^k) < \frac{2m^2}{5}.$$ As already noted above, this contradicts $$\frac{m^2}{2} < p^k.$$
Here is my question:
Does this proof argument conclusively show how to prove the Dris Conjecture that $p^k < m$? If not, how can it be mended to produce a logically sound proof?
