In what follows, denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the aliquot sum of $x$ by $s(x)=\sigma(x)-x$.
Let $m = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Using the fundamental equation $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(m)=2m=2q^k n^2$$ and using the fact that $q^k$ and $\sigma(q^k)$ are relatively prime (that is, $\gcd(q^k,\sigma(q^k))=1$), then we get that $q^k \mid \sigma(n^2)$, whereupon we obtain $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}.$$
Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ which works whenever $B \neq 0$, $D \neq 0$ and $B \neq D$, we obtain $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{(q - 1)(2n^2 - \sigma(n^2))}{q^k - 1}$$ whereupon we get, by another application of the identity above, that $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{(q - 1)(2n^2 - \sigma(n^2))}{q^k - 1}=\frac{\sigma(n^2) - (q - 1)(2n^2 - \sigma(n^2))}{q^k - (q^k - 1)},$$ from which we finally obtain $$\frac{\sigma(n^2)}{q^k}=q\sigma(n^2) - 2(q-1)n^2.$$
However, Dris (Lemma 2.1, pages 13 to 14) proved in the year $2017$ that, in fact, we have $$\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}.$$
This means that $$\gcd(n^2,\sigma(n^2))=q\sigma(n^2) - 2(q-1)n^2, \tag{1}$$ where we note that $\gcd(n^2,\sigma(n^2)) \geq 3$.
Note that Equation (1) is an instance of Bezout's Identity.
Now the hyperlinked Wikipedia page for Bezout's Identity says that the Bezout coefficients $$y = -2(q - 1)$$ and $$z = q$$ (obtained from $\gcd(n^2,\sigma(n^2))=yn^2 + z\sigma(n^2)$) are not unique.
Here then is my:
QUESTION: Can you find another pair of Bezout coefficients $(y',z')$ such that $$\gcd(n^2,\sigma(n^2))=y' n^2 + z' \sigma(n^2)$$ holds, where $y' \neq y$ and $z' \neq z$ are integers?
It appears that there is nothing specific to your numbers as far as Bezout's identity is concerned. If the equation $ax+by=\gcd(a,b)$ has a solution $(x_0,y_0)$, then all the infinitely many integer solutions to the equation are given by $x=rt+x_0,y=st+y_0$, where $r=b/\gcd(a,b)$ and $s=-a/\gcd(a,b)$.