A question related to the Descartes-Frenicle-Sorli conjecture on odd perfect numbers

Good day to everyone! I apologize in advance for the somewhat long post, but I had to put in all the details into a single question to communicate what I believe to be a viable approach to odd perfect numbers.

A positive integer $N$ is said to perfect if $\sigma(N) = 2N$, where $\sigma$ is the classical sum-of-divisors function. For example, $6$ is perfect since the divisors of $6$ are $1$, $2$, $3$ and $6$, and thus $\sigma(6) = 1 + 2 + 3 + 6 = 12 = 2 \cdot 6$. So far, all of the known examples for perfect numbers are even, and it is currently an open problem to either find an odd perfect number or show that none exists. The general consensus among mathematicians is that odd perfect numbers do not exist.

Euler gave the general form that an odd perfect number $N$ must take. This is $N = {q^k}{n^2}$ where $q$ is prime with $\gcd(q, n) = 1$ and $q \equiv k \equiv 1 \pmod 4$. Descartes (and Frenicle) believed that $k = 1$ (see Beasley's overview paper detailing this here) - more recently, Sorli conjectured the same in his Ph.D. thesis.

Lastly, Descartes "saw no reason why an odd perfect number could not exist", and gave an example of a number $D$ that comes very close to perfection (which, in modern terminology, is called a Descartes number, of which only one is currently known):

$$D = {3^2}\cdot{7^2}\cdot{11^2}\cdot{13^2}\cdot{22021} = 198585576189$$

If $22021$ were prime, then $D$ would have been an odd perfect number. Alas, $22021 = {19^2}\cdot{61}$ is composite, and the number $D$ is not perfect.

In a paper published in JIS, Dris showed that the inequality $n < q$ would imply that $k = 1$ (which we can call the Descartes-Frenicle-Sorli [or simply D-F-S] conjecture on odd perfect numbers).

In the following, assume that the D-F-S conjecture is true. Then we have $k = 1$, and the odd perfect number $\bar{N}$ takes the form $\bar{N} = qn^2$.

Since the abundancy index $I(x) = \sigma(x)/x$ is weakly multiplicative (a property which it inherits from the sum-of-divisors $\sigma(x)$), then we have:

$$I(\bar{N}) = \sigma(\bar{N})/\bar{N} = 2 = I(q)\cdot{I(n^2)}$$

because $\gcd(q, n) = 1$.

Since $\sigma(q) = q + 1 \equiv 2 \pmod 4$ and $n$ is odd, then $\sigma(q) \neq n$.

Therefore, either $\sigma(q) < n$ or $n < \sigma(q)$.

CLAIM 1: $q < n \Longleftrightarrow \sigma(q) < n$

The claim follows by observing that $q < n < \sigma(q) = q + 1$ is impossible when $q$ and $n$ are positive integers.

CLAIM 2: $\sqrt{2} < I(n)$

In fact, $$I(n) > \left(\frac{8}{5}\right)^\frac{\ln(4/3)}{\ln(13/9)} > 1.4444.$$
[This was communicated via e-mail by Ochem to Dris in April of 2013. See pages $3$ to $5$ of this paper in the arXiv for more details.]

It follows from CLAIM 2 that $I(qn) = I(q)\cdot{I(n)} > \sqrt{2}$.

We therefore have:

$$\sqrt{2} < \frac{\sigma(q)}{n}\frac{\sigma(n)}{q} = I(q)\cdot{I(n)} < 2$$

since $qn$ is deficient, being a proper divisor of the odd perfect number $\bar{N} = qn^2$.

In particular, if $q < n$, we have $\sigma(q)/n < 1$, so that:

$$\sqrt{2} < \frac{\sigma(q)}{n}\frac{\sigma(n)}{q} < \frac{\sigma(n)}{q}.$$

Thus,

$$q < n \Longrightarrow \sigma(q)/n < 1 < \sigma(q)/q \leq 6/5 < \sqrt{2} < \sigma(n)/n < \sigma(n)/q.$$

Of course, we already know that $I(n) = \sigma(n)/n < 2$ since $n$ is deficient, being a proper divisor of the odd perfect number $\bar{N} = qn^2$. Alas, I currently do not know if $\sigma(n)/q$ can be bounded from above.

By assuming the D-F-S conjecture, an obvious upper bound would be

$$\frac{\sigma(n^2)}{q} > \frac{\sigma(n)}{q}.$$

But then again, I know of no upper bound for $\sigma(n^2)/q$. If there was such an upper bound, then by the results of this paper in the arXiv, there would only be finitely many odd perfect numbers.

My question now is this: If my penultimate goal were to prove the biconditional

$$k = 1 \Longleftrightarrow n < q$$

(so that the D-F-S conjecture would then imply that the Euler prime $q$ is the largest prime factor of $\bar{N} = qn^2$), I was wondering if anyone out there have any bright ideas on how a contradiction might be derived from everything that I've written so far.

Any helpful insights will be appreciated!

Postscript: Logically, since the implication $n < q \Longrightarrow k = 1$ is true, and

since either the biconditional

$$k = 1 \Longleftrightarrow n < q$$

is true or false (but not both), then either the implication

$$k = 1 \Longrightarrow n < q$$ is true or false (but not both).

If it is false, then $k = 1$ and $q < n$, from which it follows that $q^k < n$.

On the other hand, if $n < q^k$ is true, then the biconditional

$$k = 1 \Longleftrightarrow n < q$$

is true.