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 $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
The question is as in the title:
Is it possible to improve on $I(n) > \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}$, if $\sigma(n^2)=q^k n$ and $q^k n^2$ is an odd perfect number with special prime $q$?
MY ATTEMPT
This earlier post contains a proof for the inequality $$I(n) > \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}}.$$
So now suppose that $q^k n^2$ is an odd perfect number with special prime $q$, and that $\sigma(n^2) = q^k n$. It follows that $$I(n^2) = \dfrac{q^k}{n}.$$
This implies that $$\sigma(q^k) = 2n$$ $$3 > \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} > 2 + \dfrac{1}{2}\cdot\bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \approx 2.722202786849141.$$
I now refer you to the following publication, where it is proved (on pages 106 to 107) that $$I(q^k) + I(n) \leq \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}$$ holds in general, and further on page 104 that $$I(q^k) + I(n) \geq \frac{6}{5} + \bigg(\dfrac{8}{5}\bigg)^{\dfrac{\ln(4/3)}{\ln(13/9)}} \approx 2.644405573698282.$$
MY SECOND ATTEMPT
Since $$I(n^2) = \dfrac{q^k}{n} \iff \sigma(q^k) = 2n,$$ then we obtain (per this answer to a closely related MSE question) $$3 > \dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k} = \dfrac{\sigma(q^k)}{n} + \dfrac{\dfrac{\sigma(n)}{n}}{\dfrac{q^k}{n}} = \dfrac{\sigma(q^k)}{n} + \dfrac{I(n)}{I(n^2)} \geq \dfrac{\sigma(q^k)}{n} + \prod_{p}{\dfrac{p^2 + p}{p^2 + p + 1}} = 2 + \dfrac{\zeta(3)}{\zeta(2)} \approx 2.730762969401438498726,$$ which is even higher than the previous lower bound obtained for $$\dfrac{\sigma(q^k)}{n} + \dfrac{\sigma(n)}{q^k}.$$
Hence, I am led to believe that it may in fact be possible to improve on the lower bound for $I(n)$. Alas, this is where I get stuck.
