If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m)$? (Preamble:  This question is tangentially related to this earlier post.)
Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$.  Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.  Finally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.
The topic of odd perfect numbers likely needs no introduction.
Let $p^k m^2$ be an odd perfect number with special prime $p$, satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Dris conjectured in (Dris (2008)) and (Dris (2012)) that the inequality $p^k < m$ always holds.  Brown was the first one to prove in a preprint (Brown (2016)) that the weaker inequality $p < m$ holds in general.  However, recent evidence suggests that the Dris Conjecture that $p^k < m$ may in fact be false.  (If we could rule out or show that $m < p^k$ follows from the remaining two cases in the linked question, then we will have a disproof for both the Dris Conjecture and the Descartes-Frenicle-Sorli Conjecture that $k=1$.)
Here is my question:

If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m)$?

MY ATTEMPT
Since $p$ is prime, we have
$$1<I(p^k)=\frac{\sigma(p^k)}{p^k}=\frac{p^{k+1}-1}{p^k (p - 1)}<\frac{p^{k+1}}{p^k (p - 1)}=\frac{p}{p-1}.$$
Now, because $p$ is a prime satisfying $p \equiv 1 \pmod 4$, we have the lower bound $p \geq 5$, whereupon we obtain the upper bound
$$I(p^k)<\frac{p}{p-1} \leq \frac{5}{4}.$$
Note that, since $p^k m^2$ is perfect and $\gcd(p,m)=1$, then we have
$$2=I(p^k m^2)=I(p^k)I(m^2) \iff I(m^2) = \frac{2}{I(p^k)}.$$
This implies that we have the lower bound
$$I(m^2) > \frac{2}{(5/4)} = \frac{8}{5},$$
from which we finally get
$$1 < I(p^k) < \frac{5}{4} < \frac{8}{5} < I(m^2) < 2,$$
since $m^2$ is a proper factor of the perfect number $p^k m^2$, and is therefore deficient.
But we know that
$$I(p^k) < \dfrac{5}{4} < \sqrt{\dfrac{8}{5}}< \sqrt{I(m^2)} < I(m) < I(m^2) < 2$$
In particular, we have
$$0 < 2 - I(m) < 2 - I(p^k) < 1$$
$$0 < \dfrac{D(m)}{m} < \dfrac{D(p^k)}{p^k} < 1$$
$$0 < D(m) < \frac{m}{p^k}\cdot{D(p^k)} < m.$$
We therefore have the implication
$$m < p^k \implies D(m) < D(p^k).$$
Alas, this is where I get stuck.
 A: Too long to comment :
Let $c:=\dfrac{\ln(4/3)}{\ln(13/9)}$. Using $(I(m^2))^{c} < I(m)$, one can get
$$m\le \dfrac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\implies D(m)\lt D(p^k)\tag1$$
which is better than $$m < p^k \implies D(m) < D(p^k).$$
Proof :
$(I(m^2))^c\lt I(m)$ is equivalent to
$$\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c\lt\frac{\sigma(m)}{m}\iff \sigma(m)\gt m\bigg(\frac{\sigma(m^2)}{m^2}\bigg)^c=m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
Using $\sigma(m)\gt m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$, we get
$$D(p^k)-D(m)=D(p^k)-2m+\sigma(m)\gt D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c$$
So, we can say that if
$$D(p^k)-2m+m\bigg(\dfrac{2}{I(p^k)}\bigg)^c\ge 0\tag2$$
then $D(m)\lt D(p^k)$ where note that
$$(2)\iff m\le \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^c}$$ So, we can say that $(1)$ holds.
To see that $(1)$ is better than $m < p^k \implies D(m) < D(p^k)$,  it is sufficient to prove that
$$p^k\lt \frac{D(p^k)}{2-\bigg(\dfrac{2}{I(p^k)}\bigg)^{c}}\tag3$$
We have
$$\begin{align}(3)&\iff p^k\lt \frac{p^{k+1}-2p^k+1}{(p-1)\bigg(2-\bigg(\dfrac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)}
\\\\&\iff p^k(p-1)\bigg(2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\bigg)\lt p^{k+1}-2p^k+1
\\\\&\iff 2-\bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\lt \frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt 2-\frac{p^{k+1}-2p^k+1}{p^k(p-1)}
\\\\&\iff \bigg(\frac{2p^k(p-1)}{p^{k+1}-1}\bigg)^c\gt \frac{p^{k+1}-1}{p^k(p-1)}
\\\\&\iff \frac{2p^k(p-1)}{p^{k+1}-1}\gt \bigg(\frac{p^{k+1}-1}{p^k(p-1)}\bigg)^{1/c}
\\\\&\iff 2\gt \bigg(\frac{p^{k+1}-1}{p^{k}(p-1)}\bigg)^{(c+1)/c}
\\\\&\iff 2^{c/(c+1)}\gt \frac{p^{k+1}-1}{p^{k}(p-1)}
\\\\&\iff 2^{c/(c+1)}p^k(p-1)-p^{k+1}+1\gt 0
\\\\&\iff p^k\underbrace{\bigg((2^{c/(c+1)}-1)p-2^{c/(c+1)}\bigg)}_{\text{positive}}+1\gt 0\end{align}$$
which does hold. So, $(3)$ holds.
A: Following mathlove's hint in the comments, I obtain:
Assume that
$$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} \leq m.$$
This implies that
$$\dfrac{D(p^k)}{2 - I(m^2)} \leq m$$
$$\implies \dfrac{D(p^k)}{\dfrac{D(m^2)}{m^2}} \leq m$$
$$\implies \dfrac{m^2 D(p^k)}{D(m^2)} \leq m$$
$$\implies mD(p^k) \leq D(m^2)$$
$$\implies D(p^k) \leq \dfrac{D(m^2)}{m}$$
But we obtain from
$$I(m) < I(m^2) \iff 2 - I(m^2) < 2 - I(m) \iff \dfrac{D(m^2)}{m^2} < \dfrac{D(m)}{m} \iff \dfrac{D(m^2)}{m} < D(m).$$
We conclude that
$$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} \leq m \implies D(p^k) < D(m).$$
QED
Additional Considerations
When the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds, then
$$\dfrac{D(p^k)}{2 - \dfrac{2}{I(p^k)}} = \dfrac{p - 1}{2 - \dfrac{2p}{p+1}} = \dfrac{p^2 - 1}{2}.$$
