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 $$g(5) \leq g(q) = f(k) \leq f(1)$$ 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 $$\frac{57 \times {5^{2k}} - 2 \times {5^{k+1}} + 1}{4 \times {5^k} (5^{k+1} - 1)} \leq 3 - \bigg(\frac{q - 1}{q(q + 1)}\bigg)$$ which is equivalent to $$\frac{q - 1}{q(q + 1)} \leq 3 - \Bigg(\frac{57 \times {5^{2k}} - 2 \times {5^{k+1}} + 1}{4 \times {5^k} (5^{k+1} - 1)}\Bigg) = \frac{3 \times {5^{2k}} - 2 \times {5^k} - 1}{4 \times {5^k} (5^{k+1} - 1)},$$ which, in turn, is equivalent to $$\frac{4 \times {5^k} (5^{k+1} - 1)}{3 \times {5^{2k}} - 2 \times {5^k} - 1} \leq \frac{q(q+1)}{q-1} = \frac{q(q-1)}{q-1}+\frac{2q}{q-1}=q+\frac{2q}{q-1}=q+\frac{2(q-1)}{q-1}+\frac{2}{q-1}=(q+2)+\frac{2}{q-1}.$$ But WolframAlpha says that the partial fraction decomposition of $$\frac{4 \times {5^k} (5^{k+1} - 1)}{3 \times {5^{2k}} - 2 \times {5^k} - 1}$$ is $$\frac{4 \times {5^k} (5^{k+1} - 1)}{3 \times {5^{2k}} - 2 \times {5^k} - 1} = \frac{20}{3} + \frac{4}{5^k - 1} - \frac{8}{3\bigg(3 \times {5^k} + 1\bigg)},$$ and that $$\frac{4}{5^k - 1} - \frac{8}{3\bigg(3 \times {5^k} + 1\bigg)} > 0$$ for $k \geq 1$.
Thus, we have the series of numerical inequalities $$\frac{20}{3} < \frac{4 \times {5^k} (5^{k+1} - 1)}{3 \times {5^{2k}} - 2 \times {5^k} - 1} \leq \frac{q(q+1)}{q-1} = (q+2) + \frac{2}{q-1} \leq (q+2) + \frac{2}{4} = (q+2) + \frac{1}{2} = q + \frac{5}{2}$$ $$\implies 4.1\overline{666} = \frac{25}{6} = \frac{40 - 15}{6} = \frac{20}{3} - \frac{5}{2} < q \implies q \geq 5.$$
Here are my:
QUESTIONS: Can we improve on the starting-point inequality $$g(5) \leq g(q) = f(k) \leq f(1),$$ to hopefully produce a nontrivial lower bound for $q$? If this is not possible, can you explain/show why?
Too long to comment:
From the inequality $$\frac{q-1}{q(q+1)} \leq \frac{{3 \times 5^{2k}} - {2 \times {5^k}} - 1}{4 \times {5^k}(5^{k+1} - 1)},$$ since $$h(k) := \frac{{3 \times 5^{2k}} - {2 \times {5^k}} - 1}{4 \times {5^k}(5^{k+1} - 1)}$$ is increasing for $k \geq 1$ (see this WolframAlpha computational verification for more details), then we have $$\frac{q-1}{q(q+1)} < \lim_{k \rightarrow \infty}{h(k)} = \lim_{k \rightarrow \infty}{\Bigg(\frac{{3 \times 5^{2k}} - {2 \times {5^k}} - 1}{4 \times {5^k}(5^{k+1} - 1)}\Bigg)} = \frac{3}{20},$$ which implies that $$\frac{57}{20} = 3 - \frac{3}{20} < 3 - \Bigg(\frac{q-1}{q(q+1)}\Bigg) = \frac{3q^2 + 2q + 1}{q(q + 1)} = \frac{q + 1}{q} + \frac{2q}{q + 1}.$$
But since $q$ is the special prime satisfying $q \equiv 1 \pmod 4$, then $q \geq 5$, which implies that $$I(q) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq \frac{6}{5}$$ $$I(n^2) = \frac{2}{I(q)} = \frac{2q}{q+1} \geq \frac{5}{3}.$$ Note that $$I(q) \leq \frac{6}{5} < \frac{5}{3} \leq I(n^2).$$
Consider the nonpositive product $$\Bigg(I(q) - \frac{6}{5}\Bigg)\Bigg(I(n^2) - \frac{6}{5}\Bigg) \leq 0.$$ We obtain $$I(q)I(n^2) + \bigg(\frac{6}{5}\bigg)^2 \leq \frac{6}{5}\Bigg(I(q) + I(n^2)\Bigg),$$ from which we get $$\frac{43}{15} = \frac{5}{3} + \frac{6}{5} = \frac{5}{6}\Bigg(2 + \bigg(\frac{6}{5}\bigg)^2\Bigg) \leq I(q) + I(n^2) = \frac{q+1}{q} + \frac{2q}{q+1}.$$
Notice that $$\frac{43}{15} > \frac{57}{20},$$ and that $$I(q) + I(n^2) > \frac{57}{20}$$ is trivial compared to $$I(q^k) + I(n^2) = I(q) + I(n^2) \geq \frac{43}{15},$$ where we note that the latter bound holds under the assumption $k=1$.