Hereinafter, let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer $x$ by $I(x)=\sigma(x)/x$, where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$. Note that, since $N = q^k n^2$ is perfect and $\gcd(q,n)=1$, then we have $$2=I(N)=I(q^k n^2)=I(q^k)I(n^2).$$
Inspired by the approach in this MSE question, I now consider $$f(k)=g(q):=I(q^k)+I(n^2)=I(q^k)+\frac{2}{I(q^k)}=\bigg(\frac{q^{k+1} - 1}{q^k (q - 1)}\bigg) + \bigg(\frac{2q^k (q - 1)}{q^{k+1} - 1}\bigg)=\frac{3q^{2k+2} - 4q^{2k+1} + 2q^{2k} - 2q^{k+1} + 1}{q^k (q - 1)(q^{k+1} - 1)}.$$
Note that $$\frac{\partial}{\partial k}{f(k)}=-\log(q)\Bigg[\dfrac{(((q-4)q + 2)q^k + 2q)q^k - 1}{{q^k}(q - 1)(q^{k+1} - 1)^2}\Bigg] < 0$$ (WolframAlpha computational verification is here) and that $$\frac{\partial}{\partial q}{g(q)}=\dfrac{\bigg(q(q^k - 1) - k(q - 1)\bigg)\Bigg(q^k \bigg(((q-4)q + 2)q^k + 2q\bigg) - 1\Bigg)}{{q^{k+1}}(q - 1)^2 (q^{k+1} - 1)^2} > 0$$ (WolframAlpha computational verification is here).
In particular, this means that $$g(5) \leqslant g(q) = f(k) \leqslant f(1)$$ since $g(q)$ is increasing while $f(k)$ is decreasing.
But we have $$g(5) = \frac{57 \times {5^{2k}} - 2 \times {5^{k+1}} + 1}{4 \times {5^k} (5^{k+1} - 1)}$$ while $$f(1) = \frac{3q^2 + 2q + 1}{q(q + 1)} = 3 - \bigg(\frac{q - 1}{q(q + 1)}\bigg).$$
Consequently, we obtain $$\frac{57 \times {5^{2k}} - 2 \times {5^{k+1}} + 1}{4 \times {5^k} (5^{k+1} - 1)} \leqslant 3 - \bigg(\frac{q - 1}{q(q + 1)}\bigg)$$ which is equivalent to $$\frac{q - 1}{q(q + 1)} \leqslant 3 - \Bigg(\frac{57 \times {5^{2k}} - 2 \times {5^{k+1}} + 1}{4 \times {5^k} (5^{k+1} - 1)}\Bigg) = \frac{3 \times {5^{2k}} - 2 \times {5^k} - 1}{4 \times {5^k} (5^{k+1} - 1)},$$ which, in turn, is equivalent to $$(q-1)\bigg(4 \times {5^k} (5^{k+1} - 1)\bigg) \leqslant q(q+1)\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg).$$
Setting $$N(k) := 3 \times {5^{2k}} - 2 \times {5^k} - 1$$ and $$D(k) := 4 \times {5^k} (5^{k+1} - 1)$$ gives $$(q-1)D(k) \leqslant q(q+1)N(k).$$
Note that $$D'(k) = 4 \times {5^k} (2 \times 5^{k+1} - 1) \log(5) > 0$$ and $$N'(k) = 2 \times {5^k} (3 \times 5^{k} - 1) \log(5) > 0$$ which means that $D(k)$ and $N(k)$ are increasing functions of $k$. In particular, we have the lower bounds $$D(k) \geqslant D(1) = 480 > 0,$$ and $$N(k) \geqslant N(1) = 64 > 0.$$
My problem now is how to extract a closed form for a bound on $q$ in terms of $5$ and $k$.
MY ATTEMPT
Solving the inequality $$(q-1)D(k) \leqslant q(q+1)N(k)$$ for $q$, and noting that $D(k)$ and $N(k)$ are both positive for all $k \geqslant 1$, we obtain $$qD(k) - D(k) \leqslant q^2 N(k) + qN(k)$$ $$q^2 N(k) + q(N(k) - D(k)) + D(k) \geqslant 0$$
But we have $$N(k) - D(k) = 2 \times {5^k} - 17 \times {5^{2k}} - 1 < 0.$$
We require that the discriminant $$d = (N(k) - D(k))^2 - 4N(k)D(k) \geqslant 0$$ in order for the quadratic inequality above to have solutions.
However, WolframAlpha simplifies $d$ to be $$d = \bigg(7 \times {5^{2k}} + 2 \times {5^{k+1}} - 1\bigg)^2,$$ which is indeed nonnegative. Hence, we obtain the following critical points for the quadratic inequality above:
$$q_1 := \frac{\bigg(17 \times {5^{2k}} - 2 \times {5^k} + 1\bigg) - \bigg(7 \times {5^{2k}} + 2 \times {5^{k+1}} - 1\bigg)}{2\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg)} = \frac{10 \times {5^{2k}} - 12 \times {5^k} + 2}{2\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg)}$$ and $$q_2 := \frac{\bigg(17 \times {5^{2k}} - 2 \times {5^k} + 1\bigg) + \bigg(7 \times {5^{2k}} + 2 \times {5^{k+1}} - 1\bigg)}{2\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg)} = \frac{24 \times {5^{2k}} + 8 \times {5^k}}{2\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg)}.$$ (Note that $q_1 < q_2$.)
We obtain the solutions $$\bigg(q \leqslant q_1\bigg) \lor \bigg(q \geqslant q_2\bigg).$$ Of these, the solutions for $$q \leqslant q_1$$ are extraneous, since $q$ being the special prime means that $q \geq 5$.
Letting $$h(k) := q_2 = \frac{24 \times {5^{2k}} + 8 \times {5^k}}{2\bigg(3 \times {5^{2k}} - 2 \times {5^k} - 1\bigg)}$$ and differentiating with respect to $k$, we get $$h'(k) = -\frac{\bigg(4 \times {5^k}\bigg)\log(5)}{(5^k - 1)^2} < 0$$ which means that $h(k)$ is decreasing.
Hence, we obtain $$q \geqslant q_2 = h(k) \geqslant \lim_{k \rightarrow \infty}{h(k)} = \lim_{k \rightarrow \infty}\Bigg(\frac{24 \times {5^{2k}} + 8 \times {5^k}}{2(3 \times {5^{2k}} - 2 \times {5^k} - 1)}\Bigg) = 4,$$ which does not improve on the known lower bound $q \geqslant 5$.
Here are my questions:
INQUIRIES
(1) Is the proof above for $q \geqslant 4$ logically sound?
(2) If the answer to Question (1) is NO, how can the proof be mended so as to produce a valid result?
(3) Can we do better than $q \geqslant 4$?
Added on January 11, 2022 - 1:36 PM Manila time
Note that we appear to have obtained the final resulting inequality $$q \geqslant \frac{24 \times {5^{2k}} + 8 \times {5^k}}{2(3 \times {5^{2k}} - 2 \times {5^k} - 1)} \geqslant 4.$$
This inequality is equivalent to the lower bound $$k \geqslant \log_{5}\Bigg(\frac{q}{q-4}\Bigg) \geqslant 0,$$ so that we have the implications $$k = 1 \implies q \geqslant 5$$ and $$q = 5 \implies k \geqslant 1,$$ both of which are trivial.
On OP's request, I am converting my comment into an answer.
For (1), I think that the answer is yes.
For (3), I think that what you have done suggests that using only $g(5)\leqslant f(1)$, $q\geqslant q_2\gt 4$ is the best result that you can get. In other words, I think that using only $g(5)\leqslant f(1)$, you cannot get any better result than $q\geqslant q_2\gt 4$. This does not mean that you cannot get any better result than $q\geqslant q_2\gt 4$.
An explanation for why $q_2\not=4$.
Suppose that there is $k$ such that $q_2=4$. Then, $$\begin{align}\frac{12 \times {5^{2k}} + 4 \times {5^k}}{3 \times {5^{2k}} - 2 \times {5^k} - 1}=4&\implies 12 \times {5^{2k}} + 4 \times {5^k}=12\times {5^{2k}} - 8 \times {5^k} - 4\\\\&\implies 3 \times {5^k}= - 1\end{align}$$ which is impossible.