On a curious equation regarding odd perfect numbers Let $p^k m^2$ be an odd perfect number (OPN) with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Since $\gcd(p^k, \sigma(p^k))=1$, then we essentially get the equation
$$\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)(p-1)}{p^k - 1},$$
where $\sigma(x)$ is the classical sum of divisors of the positive integer $x$, and $D(x)=2x-\sigma(x)$ is the deficiency of $x$.
We derive
$$\frac{\sigma(m^2)}{p^k}=\sigma(m^2) - D(m^2)(p-1) = \sigma(m^2) - 2(p-1)m^2 + (p-1)\sigma(m^2)$$
$$=2(1-p)m^2 + p\sigma(m^2)=\Bigg(I(m^2)-\frac{2(p-1)}{p}\Bigg)\cdot{pm^2},$$
where $I(x)=\sigma(x)/x$ is the abundancy index of $x$.
Here is my question:

Is it then possible to get a positive lower bound on the quantity
$$I(m^2)-\frac{2(p-1)}{p}?$$

Note that I know the following facts:
(1) $\sigma(m^2)/p^k \geq 315$ by a result of Broughan, Delbourgo, and Zhou.
(2) $2(p-1)/p < I(m^2) \leq 2p/(p+1)$ follows from $(p+1)/p \leq I(p^k) < p/(p-1)$
 A: On OP's request, I am converting my comment into an answer.
FYI :
$I(m^2)-\dfrac{2(p-1)}{p}$ can be written as
$$I(m^2)-\frac{2(p-1)}{p}=\frac{\sigma(m^2)}{m^2}-\frac{2(p-1)}{p}=\frac{2p^k}{\sigma(p^k)}-\frac{2(p-1)}{p}$$
$$=\frac{2p^k(p-1)}{p^{k+1}-1}-\frac{2(p-1)}{p}=\frac{2p^{k+1}(p-1)-2(p-1)(p^{k+1}-1)}{p(p^{k+1}-1)}=\frac{2(p-1)}{p(p^{k+1}-1)}$$
Let $f(k,p):=\dfrac{2(p-1)}{p(p^{k+1}-1)}$.
Then, we have $$\dfrac{\partial f(k,p)}{\partial k}\lt 0,\qquad \dfrac{\partial f(k,p)}{\partial p}\lt 0$$
and
$$\lim_{k\to\infty}f(k,p)=\lim_{p\to \infty}f(k,p)=0$$
A: This answer offers another perspective for the problem at hand.

Dris (2012) showed that
$$\frac{\sigma(m^2)}{p^k} > 1 \tag{1}$$
must necessarily hold.
But note that we have
$$\frac{\sigma(m^2)}{p^k} = \Bigg(I(m^2) - \frac{2(p-1)}{p}\Bigg)\cdot{pm^2} \tag{2}$$
Inequality $(1)$ and Equation $(2)$ together imply that
$$\Bigg(I(m^2) - \frac{2(p-1)}{p}\Bigg)\cdot{pm^2} > 1$$
so that we finally obtain
$$I(m^2) - \frac{2(p-1)}{p} > \frac{1}{pm^2} \tag{3}$$
Notice that
$$\varepsilon = \frac{1}{pm^2} > 0,$$
albeit very small.
