(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.)