(Note: A detailed version of this question was posted in MSE last April 15, 2020. It has not received any responses there as of yet. I have therefore cross-posted it here, hoping that it is suitable for this site (i.e. that it is research-level) and that it can get answered here.)
My primary aim with pursuing this question has been to resolve the problem as described in this MSE post. Note that comments below this MO question show that there are no odd solutions below ${10}^{10}$ to the closely related equation $$\sigma(\sigma(x^2))=2x\sigma(x)$$ where $\sigma=\sigma_1$ is the classical sum of divisors.
So now, let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{\gcd(n^2,\sigma(n^2))}$$ we obtain $$\gcd(\sigma(q^k),\sigma(n^2))=\gcd(n^2,\sigma(n^2))$$ if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$. This latter condition holds, for example, when $\sigma(n^2)=q^k n$.
Now the GCD $$\gcd(n^2,\sigma(n^2))$$ can be easily expressed in terms of a linear combination of $n^2$ and $\sigma(n^2)$, via $$\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2-\sigma(n^2)}{\sigma(q^{k-1})}$$ yielding $$\gcd(n^2,\sigma(n^2))=2(1-q)n^2 + q\sigma(n^2).$$
Here is my question:
How can I then derive a linear combination for $\gcd(\sigma(q^k),\sigma(n^2))$ in terms of $\sigma(q^k)$ and $\sigma(n^2)$? Note that this should always be (unconditionally) possible by Bézout's Identity.
I tried using the equation $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(m)=2m=2q^k n^2$$ but since $\sigma(q^k)$ and $\sigma(n^2)$ are on the same side of the equation, I guess this does not lead anywhere?
