(Preamble: This inquiry is an offshoot of this answer to a closely related question.)
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 show sufficient conditions 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.
We then obtain 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 a related MSE question, 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 $$\begin{cases} 2 = q^k s(q^k) t - 2(q - 1)s(q^k) \\ -1 = - (\sigma(q^k)/2) s(q^k) \cdot t + q s(q^k) \end{cases} \tag{5} $$
For simpler algebra, let
$$u = q^k s(q^k) t - 2(q - 1)s(q^k), \; \; v = -(\sigma(q^k)/2)s(q^k) \cdot t + q s(q^k) \tag{6}\label{eq1A}$$
then $(4)$ becomes
$$\begin{equation}\begin{aligned} 2n^2 - \sigma(n^2) & = u \cdot n^2 + v \cdot \sigma(n^2) \\ (2 - u)n^2 & = (1 + v)\sigma(n^2) \end{aligned}\end{equation}\tag{7}\label{eq2A}$$
From $(2)$, we have
$$\gcd\bigl(n^2,\sigma(n^2)\bigr)=\frac{D(n^2)}{s(q^k)} = d \; \; \to \; \; \gcd\left(\frac{n^2}{d},\frac{\sigma(n^2)}{d}\right) = 1 \tag{8}\label{eq3A}$$
Dividing both sides of \eqref{eq2A} by $d$ gives
$$(2 - u)\left(\frac{n^2}{d}\right) = (1 + v)\left(\frac{\sigma(n^2)}{d}\right) \tag{9}\label{eq4A}$$
Due to the right side $\gcd$ condition in \eqref{eq3A}, we next have
$$\frac{n^2}{d} \mid (1 + v) \; \; \to \; \; 1 + v = m\left(\frac{n^2}{d}\right) \tag{10}\label{eq5A}$$
for some integer $m$. Substituting this into \eqref{eq4A} and dividing both sides by $\frac{n^2}{d}$ gives
$$2 - u = m\left(\frac{\sigma(n^2)}{d}\right) \tag{11}\label{eq6A}$$
So to recap, essentially we have shown that $$\gcd(1+v,2-u)=m \tag{12}$$ where $u$ and $v$ are defined in \eqref{eq1A}.
We now claim that $m = 2 - ts(q^k)$. (Note that having $m = 0$ is equivalent to the method of equating coefficients as in $(5)$.)
We have $$1+v = 1 - (\sigma(q^k)/2)s(q^k)t + qs(q^k) = m\cdot\Bigg(\frac{n^2}{d}\Bigg)=m\cdot\Bigg(\frac{n^2}{\frac{n^2}{\sigma(q^k)/2}}\Bigg)=m\cdot(\sigma(q^k)/2) \tag{13}$$ and $$2-u = 2 - q^k s(q^k) t + 2(q - 1)s(q^k) = m\cdot\Bigg(\frac{\sigma(n^2)}{d}\Bigg)=m\cdot\Bigg(\frac{\sigma(n^2)}{\sigma(n^2)/q^k}\Bigg)=m\cdot q^k. \tag{14}$$
Multiply Equation $(13)$ by $2$, then subtract Equation $(14)$, to get $$2(1+v) - (2-u) = 2v + u = -\sigma(q^k)s(q^k)t + 2qs(q^k) + q^k s(q^k) t - 2(q - 1)s(q^k) = m\cdot{s(q^k)}. \tag{15}$$
Divide Equation $(15)$ by $s(q^k) \geq 1$, which then gives $$-\sigma(q^k)t + 2q + q^k t - 2(q - 1) = m = 2 - ts(q^k).$$
QED
Using the equation $m = 2 - ts(q^k)$, one can show that the following implications hold:
$m = 0 \implies k = 1$ - The truth of this implication is shown in this post.
$m = 1 \implies k = 1$ - Proof: Let $m = 1$, and assume to the contrary that $k \neq 1$. Then $k \geq 5$ (since $k \equiv 1 \pmod 4$ holds). But $$s(q^k) = \sigma(q^k) - q^k = 1 + q + \ldots + q^{k-1} = \frac{q^k - 1}{q - 1}$$ $$\frac{\partial}{\partial k} s(q^k) = \frac{q^k \ln q}{q - 1} > 0$$ so that $s(q^k)$ is an increasing function of $k$ (for $q \geq 5$ and $k \geq 5$). In particular, $$s(q^k) = \frac{q^k - 1}{q - 1} \geq s(q^5) = \frac{q^5 - 1}{q - 1} \geq \frac{5^5 - 1}{5 - 1} = 781.$$ Hence, we have $$1 = 2 - m = ts(q^k) \geq 781t,$$ which contradicts the fact that $t$ must be an integer, since $t > 0$ follows from $1 = ts(q^k)$ and $s(q^k) > 1$.
Here then, are our questions:
QUESTION #1: Is our proof for $m = 2 - ts(q^k)$ logically correct? If not, how can it be mended so as to produce a valid argument?
QUESTION #2: Are our proofs for the implications $m = 0 \implies k = 1$ and $m = 1 \implies k = 1$ logically correct? If possible, can they be extended to a full proof for the Descartes-Frenicle-Sorli Conjecture that $k=1$?
This is an alternative proof for $m=2-ts(q^k)$, and therefore attempts to answer QUESTION #1.
We compute the fully simplified expressions for $1+v$ and $2-u$ using WolframAlpha, and we obtain $$1+v=-\frac{(q^{k+1} - 1)(tq^k - 2q - t + 2)}{2(q-1)^2}$$ and $$2-u=-\frac{q^k (tq^k - 2q - t + 2)}{q - 1}$$ so that $$m=\gcd(1+v,2-u)=-\Bigg(\frac{tq^k - 2q - t + 2}{q - 1}\Bigg)$$ since $\gcd(q^{k+1} - 1, q^k)=1$. But $$m=-\Bigg(\frac{tq^k - 2q - t + 2}{q - 1}\Bigg)=2-t\Bigg(\frac{q^k - 1}{q - 1}\Bigg)=2-ts(q^k),$$ an equation which we know holds.