If $q^k n^2$ is an odd perfect number with special prime $q$, then $n^2 - q^k$ is not a square.

Here's a way to finish the proof without appealing to any conjecture.

If $q^k n^2$ is a perfect number with $\operatorname{gcd}(q,n)=1$, we have $$ \sigma(q^k) \sigma(n^2) = 2 q^k n^2. $$ We know that $\sigma(q^k) = (q^{k+1}-1)/(q-1)$ and you've shown that $n = (q^k + 1)/2$, so we can conclude that $$ 2(q^{k+1}-1) \sigma(n^2) = (q-1) q^k (q^k + 1)^2.\tag{$*$} $$ Consider the GCD of $q^{k+1}-1$ with the right-hand side: $$ \operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\operatorname{gcd}(q^{k+1}-1,q^k+1)^2, $$ since $q^k$ is coprime to $q^{k+1} - 1$.

Noticing that $q^{k+1} - 1$ = $q(q^k + 1) - (q + 1)$, we find $\operatorname{gcd}(q^{k+1}-1,q^k+1) = \operatorname{gcd}(q+1,q^k+1)$, which is $q+1$ because $k$ is odd.

Thus $$ \operatorname{gcd}(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)(q+1)^2. $$ Since $k\equiv 1 \pmod 4$ and you have shown $k \gt 1$, we have $k \ge 5$. If $(*)$ holds, the left-hand side of the inequality must be $q^{k+1}-1$, which is then greater than $q^5$. But the right-hand side is less than $q^4$, so this is impossible.