Note that $\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2,\sigma(n^2))$ if and only if $\gcd(n,\sigma(n^2))=\gcd(n^2,\sigma(n^2))$. I have therefore undeleted this question, in view of this closely related MSE question: When does $\gcd(m,\sigma(m^2))$ equal $\gcd(m^2,\sigma(m^2))$? What are the exceptions?.
(Cross-posting this question from MO, as I think it is about to be closed there, due to some reasons yet unknown to myself.)
This question is directly related to this earlier one ("On odd perfect numbers and a GCD - Part II") in MSE.
That question has all the details.
Now, let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.
In the hyperlinked MSE question, we have the equation $$\gcd(\sigma(q^k),\sigma(n^2))=i(q)=\gcd(n^2,\sigma(n^2))$$ where $$i(q)=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
Assume to the contrary that $k=1$. By Brown (2016), it follows that $q < n$, whence it follows that $\sigma(q^k) = \sigma(q) = q + 1 < n$.
But $i(q) = \gcd(\sigma(q^k),\sigma(n^2))$; in particular, this means that $$i(q) \mid \sigma(q^k) \implies \frac{2n^2}{\sigma(q^k)} = \frac{\sigma(n^2)}{q^k} \leq \sigma(q^k).$$ We know that $\sigma(q^k) < n$, so we have $$2n < \frac{2n^2}{\sigma(q^k)} = \frac{\sigma(n^2)}{q^k} \leq \sigma(q^k) < n,$$ resulting in a contradiction.
We therefore conclude that $k \neq 1$.
