Is this a valid proof for $I(n^2) \geq \frac{5}{3}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ is said to be perfect if $\sigma(m)=2m$.  Equivalently, $I(m)=2$.
Euler proved that an odd perfect number, if one exists, must have the form
$$m = q^k n^2$$
where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime, we have
$$\frac{q+1}{q} = I(q) \leq I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{q^{k+1} - 1}{q^k (q - 1)} < \frac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}$$
from which it follows that
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Note that we then have the lower bound
$$I(n^2) > \frac{2(q-1)}{q} \geq \frac{8}{5}$$
since $q$ is a prime satisfying $q \equiv 1 \pmod 4$.
Here is my initial question:

Can we improve the lower bound for $I(n^2)$ to
$$I(n^2) \geq \frac{5}{3}$$
using the following argument?

$$\bigg(\frac{2q}{q+1} \geq I(n^2) > \frac{5}{3}\bigg) \implies q > 5 \implies q \geq 13 \implies \bigg(I(n^2) > \frac{2(q-1)}{q} \geq \frac{24}{13} > \frac{5}{3}\bigg)$$
Thus, we have the biconditional
$$I(n^2) > \frac{5}{3} \iff q > 5.$$
Next, we have the implication
$$I(n^2) = \frac{5}{3} \implies q = 5.$$
It then suffices to prove the implication
$$q = 5 \implies I(n^2) = \frac{5}{3}$$
to finally show that
$$I(n^2) \geq \frac{5}{3},$$
since $q \geq 5$ holds.
But note that, if $q=5$, then
$$\frac{5}{3} = I(n^2) = \frac{2}{I(5^k)} = \frac{2\cdot{5^k}(5-1)}{5^{k+1}-1}$$
which implies that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
Still, notice that we have
$$k=1 \implies I(q^k) = I(q) = \frac{q+1}{q} = 1 + \frac{1}{q} \leq \frac{6}{5} \implies I(n^2) = \frac{2}{I(q^k)} = \frac{2}{I(q)} \geq \frac{2\cdot{5}}{6} = \frac{5}{3},$$
which is what we set out to prove.
Here is my final question:

Would it be possible to remove the reliance of the proof on the Descartes-Frenicle-Sorli Conjecture?

 A: I think that the answer for your initial question is yes. I've found no errors in the argument.
I think that the answer for your final question is no since under the condition that $q=5$, we see that $I(n^2)\ge \dfrac 53$ is equivalent to $k=1$ as follows :
$$\begin{align}I(n^2)\ge\frac 53&\iff \frac{8\cdot 5^k}{5^{k+1}-1}\ge\frac 53
\\\\&\iff 24\cdot 5^k\ge 5(5^{k+1}-1)
\\\\&\iff 5^k\le 5
\\\\&\iff k\le 1
\\\\&\iff k=1\end{align}$$
A: Not a complete answer, just some thoughts that recently occurred to me, which would be too long to fit in the Comments section.
Since the biconditionals
$$I(n^2) > \frac{5}{3} \iff q > 5$$
and
$$I(n^2) = \frac{5}{3} \iff \bigg(q = 5 \land k = 1\bigg)$$
hold, it remains to consider what happens to the bounds for $I(n^2)$ when $q = 5$ and $k > 1$.
Since $k > 1$ and $k \equiv 1 \pmod 4$, then $k \geq 5$.  By assumption, we have $q=5$, so that we obtain
$$I(q^k) = I(5^k) \geq I(5^5) \iff I(n^2) = \frac{2}{I(q^k)} \leq \frac{2}{I(5^5)} = \frac{3125}{1953} \approx 1.6001.$$
(WolframAlpha computation for $\dfrac{2}{I(5^5)}$ is here.)
On the other hand, we have the lower bound
$$\frac{8}{5} = \frac{2\cdot(5 - 1)}{5} = \frac{2(q - 1)}{q} < I(n^2),$$
whence there is no contradiction.
It is natural then, to attempt to derive a better lower bound than
$$1.6 = \frac{8}{5} < I(n^2),$$
specifically when $q=5$ and $k>1$.

It turns out that we can do better under the case $q=5$ and $k>1$.
Cohen and Sorli ruled out $5^5$ as a possible Eulerian component $q^k$ for an odd perfect number in page 4 of their paper titled On Odd Perfect Numbers and Even 3-Perfect Numbers.
Thus, under the assumption $q=5$ and $k>1$, we have that $k \geq 9$ (since $k \equiv 1 \pmod 4$), whereupon we get
$$1.249999872 = \frac{2441406}{1953125} = \frac{5^{10} - 1}{5^9 (5 - 1)} = I(5^9) \leq I(q^k) < \frac{5}{4} = 1.25$$
$$1.6 = \frac{8}{5} < I(n^2) \leq \frac{1953125}{1220703} \approx 1.60000016384.$$
