Problem Statement
Prove the following proposition.
If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.
Motivation
Let $q^k n^2$ be an odd perfect number with special prime $q$. Then $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
By Pomerance, et al., we know that $q^k < n^2$, so that $n^2 - q^k$ is a positive integer. Also, since $n^2$ is a square and $q \equiv 1 \pmod 4$, then $$n^2 - q^k \equiv 1 - 1 \equiv 0 \pmod 4.$$
My Attempt
Suppose that $q^k n^2$ is an odd perfect number with special prime $q$, and that $n^2 - q^k = s^2$, for some $s \geq 2$.
Then $$n^2 - s^2 = q^k = (n + s)(n - s)$$ so that we obtain $$\begin{cases} {q^{k-v} = n + s \\ q^v = n - s} \end{cases}$$ where $v$ is a positive integer satisfying $0 \leq v \leq (k-1)/2$. It follows that we have the system $$\begin{cases} {q^{k-v} + q^v = q^v (q^{k-2v} + 1) = 2n \\ q^{k-v} - q^v = q^v (q^{k-2v} - 1) = 2s} \end{cases}$$
Since $q$ is a prime satisfying $q \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, from the first equation it follows that $v=0$, so that we obtain $$\begin{cases} {q^k + 1 = 2n \\ q^k - 1 = 2s} \end{cases}$$ which yields $$n = \frac{q^k + 1}{2} < q^k.$$ Lastly, note that the inequality $q<n$ has been proved by Brown (2016), Dris (2017), and Starni (2018), so that we are faced with the inequality $$q < n < q^k.$$ This implies that $k>1$.
Finally, notice that $k>1$ contradicts the Descartes-Frenicle-Sorli Conjecture, while $n<q^k$ contradicts the Dris Conjecture.
Question
Is it possible to remove the reliance of this proof on the truth of either the Descartes-Frenicle-Sorli Conjecture or the Dris Conjecture?