Let $N$ be an odd perfect number given in the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what follows, let us keep in mind the following lemma:
LEMMA: If $q^k n^2$ is an odd perfect number given in Eulerian form, then $k = 1$ holds if and only if $$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)},$$ where $s(x)=\sigma(x)-x$ is the aliquot sum of the positive integer $x$, $D(x)=2x-\sigma(x)$ is the deficiency of $x$, and $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of $x$.
Now, we start with
$$\gcd(n^2,\sigma(n^2)) = \left(q^k t - 2(q - 1)\right)n^2 + \left((-\sigma(q^k)/2)\cdot{t} + q\right)\sigma(n^2),$$
like Tony Kuria Kimani did in this recent ResearchGate preprint.
But the equation $$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2} = \frac{s(n^2)}{D(q^k)/2} = \frac{D(n^2)}{s(q^k)}$$ holds. Consequently, we obtain the simultaneous equations $$s(n^2) = (D(q^k)/2)\cdot\gcd(n^2,\sigma(n^2))$$ $$= \left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$ $$D(n^2) = {s(q^k)}\cdot\gcd(n^2,\sigma(n^2)) = \left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2).$$
We now test whether the equation $$s(n^2) = \left(\frac{q-1}{2}\right)\cdot{D(n^2)}$$ holds; that is, whether the equation $$\left(q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)\right)n^2 + \left((-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)\right)\sigma(n^2)$$ $$= \left(\frac{q-1}{2}\right)\cdot\Bigg(\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)n^2 + \left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right)\sigma(n^2)\Bigg) \tag{1}$$ is true.
For simpler algebra, let $$u_1 = q^k (D(q^k)/2) t - 2(q - 1)(D(q^k)/2)$$ $$u_2 = \left(\frac{q-1}{2}\right)\cdot\left(q^k s(q^k) t - 2(q - 1)s(q^k)\right)$$ $$v_1 = (-\sigma(q^k)/2)(D(q^k)/2)\cdot{t} + q(D(q^k)/2)$$ and $$v_2 = \left(\frac{q-1}{2}\right)\cdot\left((-\sigma(q^k)/2)s(q^k)\cdot{t} + qs(q^k)\right).$$
Then Equation $(1)$ becomes $$\left(u_1 - u_2\right)\cdot{n^2} = \left(v_2 - v_1\right)\cdot{\sigma(n^2)}. \tag{2}$$
We now attempt to get simplified expressions for $u_1 - u_2$ and $v_2 - v_1$ using WolframAlpha. We obtain the following: $$u_1 - u_2 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(tq^k - 2q + 2\right)$$ $$v_2 - v_1 = -\left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$
Dividing both sides of Equation $(2)$ by $d = \gcd(n^2,\sigma(n^2))$, we obtain $$\left(u_1 - u_2\right)\cdot\Bigg({\frac{n^2}{d}}\Bigg) = \left(v_2 - v_1\right)\cdot\Bigg(\frac{\sigma(n^2)}{d}\Bigg).$$ Recall that $$d = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$ Consequently, we have $$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = \left(v_2 - v_1\right)\cdot{q^k}. \tag{4}$$
Since $\gcd(q^k,\sigma(q^k)/2)=1$, then there exists an integer $G$ such that $$v_2 - v_1 = G\cdot\left(\sigma(q^k)/2\right).$$ Substituting into Equation $(4)$, we get $$\left(u_1 - u_2\right)\cdot\Bigg(\frac{\sigma(q^k)}{2}\Bigg) = G\cdot\left(\sigma(q^k)/2\right)\cdot{q^k},$$ from which we finally obtain $$u_1 - u_2 = G\cdot{q^k}.$$
We therefore finally get $$\gcd(u_1 - u_2, v_2 - v_1) = \gcd\Bigg(G\cdot{q^k}, G\cdot\left(\sigma(q^k)/2\right)\Bigg) = G\cdot\gcd(q^k, \sigma(q^k)/2) = G. \tag{5}$$
To recap, we have obtained $$G = \gcd(u_1 - u_2, v_2 - v_1).$$
However, from the equations in $(3)$, we also get $$\gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$
Hence, we finally have $$G = \gcd(u_1 - u_2, v_2 - v_1) = -\left(\frac{q^k - q}{2(q - 1)}\right).$$
Since GCDs are always nonnegative, then $G \geq 0$, which means that we have $$q^k - q \leq 0.$$ This implies that $k \leq 1$. Since $k \geq 1$ ought to hold (because $k$ is a positive integer satisfying $k \equiv 1 \pmod 4$), then we now know that $k=1$.
Quite apparently, this implies that $G = 0$, whereupon we obtain $u_1 = u_2$ and $v_1 = v_2$.
Here is my question:
Can I define GCDs to be always positive (so that $G > 0$ in the previous section) and thereby get a proof for $k \neq 1$?
Posting this self-answer, thanks to hints from Professor Dujella in the chat room.
First, we show that the assumption $G > 0$ leads to a contradiction.
It is clear that, under the assumption $G > 0$, $k \neq 1$ holds. (This is because $k = 1$ implies that $G = 0$.)
Then I obtain (from Equation $(4)$ in the original post) that $$I(q^k) = \frac{\sigma(q^k)}{q^k} = \frac{2(v_2 - v_1)}{u_1 - u_2} = \frac{q(tq^k - 2q + 2) - t}{(q - 1)(tq^k - 2q + 2)} = \frac{q}{q - 1} - \frac{t}{(q - 1)(tq^k - 2q + 2)}. \tag{1}$$
(Note that from Equation $(1)$ in this answer, we have $t \neq 0$.)
We obtain $$\frac{1}{q^k (q - 1)} = \frac{t}{(q - 1)(tq^k - 2q + 2)}$$
which is equivalent to $$0 = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{1}{q^k} - \frac{t}{tq^k - 2q + 2}\right) = \left(\frac{1}{q - 1}\right)\cdot\left(\frac{tq^k - 2q + 2 - tq^k}{tq^k - 2q + 2}\right) = -\frac{2(q - 1)}{(q - 1)(tq^k - 2q + 2)}.$$ Now, since a priori we know that $q \geq 5$ holds, then we may cancel $q - 1$ from both numerator and denominator of the last fraction, to get $$0 = -\frac{2}{tq^k - 2q + 2},$$ resulting in the contradiction $$0 = -2.$$
(Note that $tq^k - 2q + 2 \neq 0$, because otherwise we would get $$t = \frac{2(q - 1)}{q^k},$$ which implies that $q^k \mid 2(q - 1)$ (since $t$ must be an integer), contradicting $\gcd(2,q)=\gcd(q-1,q)=1$.)
Hence, we conclude that $G = 0$ must hold.
Now, we claim that:
Proof: The implication $k = 1 \implies G = 0$ is obvious.
For the converse, assume that $G = 0$.
Then we obtain the system of equations $$0 = u_1 - u_2 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot(tq^k - 2q + 2) \tag{2}$$ $$0 = v_2 - v_1 = - \left(\frac{q^k - q}{2(q - 1)}\right)\cdot\left(\frac{q(tq^k - 2q + 2) - t}{2(q - 1)}\right). \tag{3}$$
Now, note that (referencing Equations $(2)$ and $(3)$ in this answer), we have $$(2) \iff \left((q^k - q = 0) \lor (tq^k - 2q + 2 = 0)\right) \land (q \neq 1) \iff k = 1,$$ since $tq^k - 2q + 2 = 0$ contradicts the requirement that $t$ must be an integer. $$(3) \iff \left((q^k - q = 0) \lor (q(tq^k - 2q + 2) - t = 0)\right) \land (q \neq 1) \iff \left((k = 1) \lor (t(q^{k+1} - 1) = 2q(q - 1))\right) \land (q \neq 1) \iff \Bigg((k = 1) \lor \left(1 < t = \frac{2q}{\sigma(q^k)} = \frac{2q}{q + 1} < 2\right)\Bigg) \land (q \neq 1) \iff k = 1,$$ which completes the proof.