Can this bound for the abundancy index of $n$ be improved, given that $q^k n^2$ is an odd perfect number with $k=1$?

(Not a complete answer.)

Let $q^k n^2$ be an odd perfect number with special/Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

First off, we show the following lemma.

Lemma 1 $I(n) \neq 3/2 = I(2)$

Proof The proof is trivial and follows from the fact that $n \neq 2$, since $n$ must be odd. (In fact, we know by Nielsen's latest result that $\omega(n) \geq 9$, where $\omega(n)$ is the number of distinct prime factors of $n$. In particular, this means that $n$ is composite.) Note that $2$ is solitary as it is prime.

Next, we prove the following implication.

Theorem $\bigg((k=1) \land (I(n) < 3/2)\bigg) \implies q=5$

Proof Since $k=1$, then we obtain $$\frac{2q}{q+1}=I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$ Using WolframAlpha, we get that $-1 < q < 5.23316$, from which we conclude that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.

In general, we can remove the assumption that $k=1$:

Theorem $I(n) < 3/2 \implies q=5$

Proof We proceed as follows: $$\frac{2(q-1)}{q} < I(n^2) \leq (I(n))^{\frac{\ln(13/9)}{\ln(4/3)}} < \bigg(\frac{3}{2}\bigg)^{\frac{\ln(13/9)}{\ln(4/3)}}.$$

Using WolframAlpha, we obtain $0 < q < 6.23316$, which implies that $q=5$, since $q$ is a prime number satisfying $q \equiv 1 \pmod 4$.

Remarks: Note that $I(n) > 3/2$ would also follow from the inequality $D(n) < s(n)$, where $D(n)=2n-\sigma(n)$ is the deficiency of $n$ and $s(n)=\sigma(n) - n$ is the sum of the aliquot divisors of $n$.