Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the GCDs $$G = \gcd(\sigma(q^k),\sigma(n^2))$$ $$H = \gcd(n^2,\sigma(n^2))$$ and $$I = \gcd(n,\sigma(n^2)),$$ where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.
Note that $$G \times H = I^2$$ $$G \mid I$$ $$I \mid H$$ and that $$H = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
Let $$J = \frac{I}{G} = \frac{H}{I} = \sqrt{\frac{H}{G}}.$$
It can be shown that $$J = \frac{n}{\gcd(\sigma(q^k)/2, n)}.$$
Suppose that $$J = \frac{n}{\gcd(\sigma(q^k)/2, n)} = \frac{n}{d}, \tag{1}$$ where $d$ is a divisor of $n$.
Note that $(1)$ is true if and only if $\gcd(\sigma(q^k)/2, n) = d$.
Notice that $$\frac{Jnd}{\sigma(q^k)/2} = \frac{n^2}{\sigma(q^k)/2} = H.$$
But we also know that $$H = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
Hence, under the given assumption, we get $$\sigma(n^2) = \frac{Jnd}{\sigma(q^k)/2}\cdot{q^k}$$ $$n = Jd.$$
Consequently, we obtain $$\gcd(n,\sigma(n^2)) = \gcd\left(Jd,\frac{Jnd}{\sigma(q^k)/2}\cdot{q^k}\right)$$ Since $\gcd(\sigma(q^k)/2, n) = d$, then $d \mid \sigma(q^k)/2$ holds.
Since $\gcd(q^k,\sigma(q^k)/2)=1$, we consider three cases:
Case (1): $\sigma(q^k)/2 \mid d$
This case implies that $d = \sigma(q^k)/2$, and therefore that $\sigma(q^k)/2 \mid n$ holds.
Case (2): $\sigma(q^k)/2 \mid n$
No problem.
Case (3): $\sigma(q^k)/2 \mid J$
$$\sigma(q^k)/2 \mid J \implies J = e\cdot\sigma(q^k)/2 \text{ for some } e \implies \frac{n}{\gcd(\sigma(q^k)/2, n)} = e\cdot\sigma(q^k)/2$$ $$\implies n = e\cdot{\sigma(q^k)/2}\cdot{\gcd(\sigma(q^k)/2, n)} \implies \sigma(q^k)/2 \mid n$$
Here is my:
QUESTION: Does this proof suffice? I have doubts if the list of three cases is complete, as I am aware that it may happen that either $\sigma(q^k)/2 \mid Jn$, $\sigma(q^k)/2 \mid Jd$, or $\sigma(q^k)/2 \mid nd$ also hold.
Update (Added on March 26, 2023 - 15:22 PM (Manila time))
The condition $\sigma(q^k)/2 \mid Jd$ is equivalent to $\sigma(q^k)/2 \mid n$ since $Jd = n$.
That leaves us (???) with the following three further cases to consider: $\sigma(q^k)/2 \mid Jn$, $\sigma(q^k)/2 \mid nd$, or $\sigma(q^k)/2 \mid Jnd$.
Alas, this is where I get stuck!