(Preamble: This question is an offshoot of this earlier post.)
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 $N = q^k n^2$ 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 this proof-verification request, we shall attempt to prove this conjecture.)
Using the fundamental equation $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2$$ 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 that $$\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}$$ is an odd integer.
From the hyperlinked MSE question above, we have obtained the identities $$\gcd\bigl(n^2,\sigma(n^2)\bigr)=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{s(q^k)}=\frac{2s(n^2)}{D(q^k)}.$$
Now, in a comment underneath MSE user Aravind's answer to the hyperlinked MSE question above, we derived that $$\gcd\bigl(n^2,\sigma(n^2)\bigr)=n^2 \bigl(q^k t - 2(q - 1)\bigr) + \sigma(n^2) \bigl( - (\sigma(q^k)/2)\cdot t + q \bigr), \tag{1}$$ where $t$ is necessarily an integer to be determined by using the identity $$\gcd\bigl(n^2,\sigma(n^2)\bigr)=\frac{D(n^2)}{s(q^k)}. \tag{2}$$
Equating (1) and (2), we obtain $$\frac{D(n^2)}{s(q^k)}=n^2 \bigl(q^k t - 2(q - 1)\bigr) + \sigma(n^2) \bigl( - (\sigma(q^k)/2)\cdot t + q \bigr). \tag{3}$$ Multiplying both sides of (3) by $s(q^k)$, we obtain $$2n^2 - \sigma(n^2) = D(n^2) = n^2 \bigl(q^k s(q^k) t - 2(q - 1)s(q^k)\bigr) + \sigma(n^2) \bigl( - (\sigma(q^k)/2) s(q^k) \cdot t + q s(q^k) \bigr). \tag{4}$$
Equating coefficients for $n^2$ and $\sigma(n^2)$ in (4), we have that $$\begin{cases} 2 = q^k s(q^k) t - 2(q - 1)s(q^k) \\ -1 = - (\sigma(q^k)/2) s(q^k) \cdot t + q s(q^k) \end{cases} \tag{5} $$
The system in (5) is a set of two simultaneous "linear" equations in $t$, of the form $$\begin{cases} A t + B = 2 \\ C t + D = -1 \end{cases} \tag{6} $$ where $$A = q^k s(q^k), B = - 2(q - 1)s(q^k)$$ and $$C = - (\sigma(q^k)/2) s(q^k), D = q s(q^k).$$
We now solve the system (6) for $t$. We thereby get
$$\begin{cases}
2D = ADt + BD \\
B = -BCt - BD
\end{cases} \tag{7}
$$
Adding the two equations in (7), we obtain
$$2D + B = (AD - BC)t$$
$$t = \frac{2D + B}{AD - BC}.$$
But $$2D + B = 2qs(q^k) - 2(q - 1)s(q^k) = s(q^k)\bigl(2q - 2(q - 1)\bigr) = 2s(q^k) \tag{8}$$ and $$AD - BC = q^{k+1} \bigl(s(q^k)\bigr)^2 - (q - 1)\sigma(q^k) \bigl(s(q^k)\bigr)^2 = \bigl(s(q^k)\bigr)^2 \bigl(q^{k+1} - (q - 1)\sigma(q^k)\bigr) $$ $$= \bigl(s(q^k)\bigr)^2 \bigl(q^{k+1} - (q^{k+1} - 1)\bigr) = \bigl(s(q^k)\bigr)^2. \tag{9}$$
Consequently, from (8) and (9), the variable $t$ (which is necessarily an integer) must have the value $$t = \frac{2D + B}{AD - BC} = \frac{2s(q^k)}{\bigl(s(q^k)\bigr)^2} = \frac{2}{s(q^k)}.$$
Since $s(q^k) \geq 1$ holds in general, then $t$ is an integer implies that either $s(q^k) = 1$ or $s(q^k) = 2$.
Want to show: $s(q^k) = 1$ Assume to the contrary that $s(q^k) = 2$. This is equivalent to $$q^k - 1 = 2(q - 1)$$ $$q^k - 2q = -1$$ $$2q - q^k = 1$$ $$q\bigl(2 - q^{k-1}\bigr) = 1$$ This implies that $q = 1$ AND $2 - q^{k - 1} = 1$, from which we obtain $1 = q^{k - 1}$ and $q = 1$. The first equation implies that $k = 1$, while the second condition contradicts $q$ being the special prime satisfying $q \equiv 1 \pmod 4$ (so that $q \geq 5$).
Hence, necessarily we have $s(q^k) = 1$, or $k = 1$.
Here is my question:
Is the method of equating coefficients for $n^2$ and $\sigma(n^2)$ in (5) a valid approach?
For simpler algebra, let
$$u = q^k s(q^k) t - 2(q - 1)s(q^k), \; \; v = -(\sigma(q^k)/2)s(q^k) \cdot t + q s(q^k) \tag{1}\label{eq1A}$$
then your $(4)$ becomes
$$\begin{equation}\begin{aligned} 2n^2 - \sigma(n^2) & = u \cdot n^2 + v \cdot \sigma(n^2) \\ (2 - u)n^2 & = (1 + v)\sigma(n^2) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
From your $(2)$, we have
$$\gcd\bigl(n^2,\sigma(n^2)\bigr)=\frac{D(n^2)}{s(q^k)} = d \; \; \to \; \; \gcd\left(\frac{n^2}{d},\frac{\sigma(n^2)}{d}\right) = 1 \tag{3}\label{eq3A}$$
Dividing both sides of \eqref{eq2A} by $d$ gives
$$(2 - u)\left(\frac{n^2}{d}\right) = (1 + v)\left(\frac{\sigma(n^2)}{d}\right) \tag{4}\label{eq4A}$$
Due to the right side $\gcd$ condition in \eqref{eq3A}, we next have
$$\frac{n^2}{d} \mid (1 + v) \; \; \to \; \; 1 + v = m\left(\frac{n^2}{d}\right) \tag{5}\label{eq5A}$$
for some integer $m$. Substituting this into \eqref{eq4A} and dividing both sides by $\frac{n^2}{d}$ gives
$$2 - u = m\left(\frac{\sigma(n^2)}{d}\right) \tag{6}\label{eq6A}$$
Regarding your question of
This is equivalent to having $m = 0$ in \eqref{eq5A} and \eqref{eq6A}. However, it seems you just stated this, without providing anything in your derivation that requires this to be the case. Thus, unless I'm missing something (if so, please indicate what that is), I believe you haven't provided sufficient proof to show that restriction (i.e., your method of equating coefficients) is valid, which means that $m$ can possibly be some non-zero integer instead.