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(This is not a direct answer to the original question, as posed, but rather are some thoughts that recently occurred to me, which would be too long to fit in the Comments section.)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Since $m^2 - p^k \neq \square$, then $p^k \neq 2m - 1$.

If $m \leq \sigma(p^k)/2$ and $p^k < 2m - 1$ both hold, then $p < 2m - 1$ is true. Note that $p^k < 2m - 1$ by itself implies $p < 2m - 1$. This means that $m \leq \sigma(p^k)/2$ cannot hold, under the case $p^k < 2m - 1$. Therefore, $\sigma(p^k)/2 < m$ must be true, under the assumption $p^k < 2m - 1$. (Note that this then implies that $p \leq p^k < 2m - 1 < 2m$.)

Now assume that $2m - 1 < p^k$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies $m < p \iff k = 1$. Substituting $k = 1$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $2m - 1 < p^k$, that $$p < 2m - 1.$$ In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)

(This is not a direct answer to the original question, as posed, but rather are some thoughts that recently occurred to me, which would be too long to fit in the Comments section.)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Since $m^2 - p^k \neq \square$, then $p^k \neq 2m - 1$.

If $p^k < 2m - 1$ holds, then $p < 2m - 1$ is true. (In particular, note that we get $p \leq p^k < 2m - 1 < 2m$.)

Now assume that $2m - 1 < p^k$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies that the biconditional $m < p \iff k = 1$ holds. Substituting $k = 1$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $2m - 1 < p^k$, that $$p < 2m - 1.$$ In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Since $m^2 - p^k \neq \square$, then $p^k \neq 2m - 1$.

If $m \leq \sigma(p^k)/2$ and $p^k < 2m - 1$ both hold, then $p < 2m - 1$ is true. Note that $p^k < 2m - 1$ by itself implies $p < 2m - 1$. This means that $m \leq \sigma(p^k)/2$ cannot hold, under the case $p^k < 2m - 1$. Therefore, $\sigma(p^k)/2 < m$ must be true, under the assumption $p^k < 2m - 1$. (Note that this then implies that $p \leq p^k < 2m - 1 < 2m$.)

Now assume that $2m - 1 < p^k$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies $m < p \iff k = 1$. Substituting $k = 1$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $2m - 1 < p^k$, that $$p < 2m - 1.$$ In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)

(This is not a direct answer to the original question, as posed, but rather are some thoughts that recently occurred to me, which would be too long to fit in the Comments section.)

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Since $m^2 - p^k \neq \square$, then $p^k \neq 2m - 1$.

If $m \leq \sigma(p^k)/2$ and $p^k < 2m - 1$ both hold, then $p < 2m - 1$ is true. Note that $p^k < 2m - 1$ by itself implies $p < 2m - 1$. This means that $m \leq \sigma(p^k)/2$ cannot hold, under the case $p^k < 2m - 1$. Therefore, $\sigma(p^k)/2 < m$ must be true, under the assumption $p^k < 2m - 1$. (Note that this then implies that $p \leq p^k < 2m - 1 < 2m$.)

Now assume that $2m - 1 < p^k$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies $m < p \iff k = 1$. Substituting $k = 1$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $2m - 1 < p^k$, that $$p < 2m - 1.$$ In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)

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 $p^k > 2m - 1$.

Let $p^k m^2$ be an odd perfect number with special prime $p$.

Since $m^2 - p^k \neq \square$, then $p^k \neq 2m - 1$.

If $m \leq \sigma(p^k)/2$ and $p^k < 2m - 1$ both hold, then $p < 2m - 1$ is true. Note that $p^k < 2m - 1$ by itself implies $p < 2m - 1$. This means that $m \leq \sigma(p^k)/2$ cannot hold, under the case $p^k < 2m - 1$. Therefore, $\sigma(p^k)/2 < m$ must be true, under the assumption $p^k < 2m - 1$. (Note that this then implies that $p \leq p^k < 2m - 1 < 2m$.)

Now assume that $2m - 1 < p^k$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $\sigma(p^k) > 2m$

Proof: $\sigma(p^k) \geq p^k + 1 > 2m$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $2pm - 2m = 2(p - 1)m = 2pm - p - 1$, which is equivalent to $p = 2m - 1 > m$. This implies that $k = 1$. But then $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $p > 2m - 1 > m$, which is equivalent to $k = 1$, since the assumption $2m - 1 < p^k$ implies $m < p^k$, which in turn implies $m < p \iff k = 1$. Substituting $k = 1$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $2m - 1 < p^k$, that $$p < 2m - 1.$$ In particular, we obtain $k \neq 1$. Since the assumption $2m - 1 < p^k$ implies $m < p^k$, and because $m < p^k$ implies the biconditional $m < p \iff k = 1$ holds, then $k \neq 1$ is equivalent to $p < m$.

Either way, we conclude that $p < 2m$. (Note that Acquaah and Konyagin has already proved that $p < m\sqrt{3}$, so all of this is but an academic exercise.)