Is it possible to prove $I(m^2) > \zeta(2) \approx 1.644934$, if $p^k m^2$ is an odd perfect number with special prime $p$? The topic of odd perfect numbers likely needs no introduction.
Let $\sigma=\sigma_{1}$ denote the classical sum of divisors.  Denote the abundancy index by $I(x)=\sigma(x)/x$.
An odd perfect number $N$ is said to be given in Eulerian form if
$$N = p^k m^2$$
where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
The question is as in the title:

Is it possible to prove $I(m^2) > \zeta(2) \approx 1.644934$, if $p^k m^2$ is an odd perfect number with special prime $p$?

MY ATTEMPT
By basic considerations, since $p$ is the special prime and satisfies $p \equiv 1 \pmod 4$, then $p \geq 5$ holds, from which it follows that
$$I(p^k)=\dfrac{\sigma(p^k)}{p^k}=\dfrac{p^{k+1}-1}{p^k (p-1)}<\dfrac{p^{k+1}}{p^k (p-1)}=\dfrac{p}{p-1} \leq \frac{5}{4} \iff I(m^2)=\frac{2}{I(p^k)}>\dfrac{2(p-1)}{p} \geq \frac{8}{5}.$$
Now, I was thinking of attempting to improve this trivial lower bound to
$$I(m^2) > \zeta(2) \approx 1.644934.$$
But I know that
$$\zeta(2) = \prod_{\rho}{\bigg({\rho}^2 + {\rho} + 1\bigg)},$$
where $\rho$ runs over all primes.  (I am not really too sure though, if that is really how I should define $\zeta(2)$.  Any way, I just based my definition off this answer to a closely related MSE question.)
Update (September 18, 2020 - 6:16 PM Manila time) I was wrong, the correct formula for $\zeta(2)$ should have been
$$\zeta(2) = \prod_{\rho}{\dfrac{{\rho}^2}{(\rho - 1)(\rho + 1)}},$$
as correctly pointed out by mathlove.
Note that we can write
$$m = \prod_{i=1}^{\omega(m)}{{\rho_i}^{\alpha_i}}$$
so that we have
$$m^2 = \prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}$$
and therefore
$$\sigma(m^2) = \sigma\Bigg(\prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}\Bigg) = \prod_{i=1}^{\omega(m)}{\sigma\bigg({\rho_i}^{2\alpha_i}\bigg)}$$
from which we get
$$I(m^2) = \dfrac{\displaystyle\prod_{i=1}^{\omega(m)}{\sigma\bigg({\rho_i}^{2\alpha_i}\bigg)}}{\displaystyle\prod_{i=1}^{\omega(m)}{{\rho_i}^{2\alpha_i}}}.$$
This is where I get stuck.  I currently do not see a way to force the inequality
$$I(m^2) > \prod_{\rho}{\bigg({\rho}^2 + {\rho} + 1\bigg)},$$
where $\rho$ runs over all primes, from everything that I have written so far.
 A: As noted, if $x=p^km^2$ is an odd perfect number with special prime $p$, so that $p\equiv k\equiv1\pmod{4}$ and $\gcd(p,m)=1$, it follows that
$$I(m^2)=\frac{\sigma(m^2)}{m^2}=\frac{p^k}{\sigma(p^k)}\frac{\sigma(p^k)}{p^k}\frac{\sigma(m^2)}{m^2}=\frac{p^k}{\sigma(p^k)}\frac{\sigma(p^km^2)}{p^km^2}=\frac{p^k}{\sigma(p^k)}I(x)=2\frac{p^k}{\sigma(p^k)},$$
where of course
$$\sigma(p^k)=\sum_{i=0}^kp^i=\frac{p^{k+1}-1}{p-1},$$
from which it follows that
$$I(m^2)=2\frac{p^k(p-1)}{p^{k+1}-1}=2\frac{p-1}{p-\tfrac{1}{p^k}}.$$
The latter allows several simple lower bounds. For example
$$2\frac{p-1}{p-\tfrac{1}{p^k}}>2\frac{p-1}{p}=2\left(1-\frac1p\right)\geq\frac85,$$
because $p\geq5$. In particular, if $p\neq5$ we see that $p\geq13$ and so
$$I(m^2)>2\left(1-\frac1p\right)\geq2\left(1-\frac{1}{13}\right)=\frac{24}{13}>\zeta(2).$$
So it remains to show that the inequality holds if $p=5$. In this case we have
$$I(m^2)=2\frac{p-1}{p-\tfrac{1}{p^k}}=\frac{8}{5-\tfrac{1}{5^k}}.$$
This is a strictly decreasing function of $k$, and for $k=1,5$ we find that
$$I(m^2)=\frac{8}{5-\tfrac15}=\frac{5}{3}>\zeta(2),$$
$$I(m^2)=\frac{8}{5-\tfrac{1}{5^5}}=\frac{3125}{1953}<\zeta(2),$$
so your question is equivalent to asking whether there exists an odd perfect number $x$ of the form $x=5^km^2$ with $5\nmid m$ and $k\equiv1\pmod{4}$ and $k\geq5$.
