Let $N = p^k m^2$ be an odd perfect number with special / Euler prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m)=1$.
First, we state the following conjectures, which are currently open (as of January 2020):
Descartes-Frenicle-Sorli Conjecture
If $N = p^k m^2$ is an odd perfect number with special / Euler prime $p$, then $k = 1$ always holds.
Dris Conjecture
If $N = p^k m^2$ is an odd perfect number with special / Euler prime $p$, then $p^k < m$ always holds.
Remark 1: Note that Dris Conjecture implies that $p < m$.
Motivation for the Present Blog Post
Since $m$ is odd, $m^2 \equiv 1 \pmod 4$. Additionally, $p \equiv k \equiv 1 \pmod 4$ imply that $p^k \equiv 1 \pmod 4$. Together, the congruences for $m^2$ and $p^k$ imply that $m^2 - p^k \equiv 0 \pmod 4$. Additionally, note that it is known that
$$p^k < \frac{2}{3}{m^2},$$
so that we know a priori that
$$m^2 - p^k > \frac{p^k}{2}.$$
In particular, we are sure that
$$m^2 - p^k > 0.$$
Since $m^2 - p^k > 0$ and $m^2 - p^k \equiv 0 \pmod 4$, we are led to the considerations in the following MSE question:
We reproduce here the accepted answer by MSE user mathlove:
This is a partial answer.
Your idea works for $z=2^{2n}$.
Claim: For $z=2^{2n}$, if $2^{n+2}+1$ is prime, then $(m,p,k)=(2^{n+1}+1,2^{n+2}+1,1)$. If $2^{n+2}+1$ is not prime, then there is no solution.
Proof:
For $z=2^{2n}$, we have
$$m^2 - p^k = 2^{2n+2}\iff p^k=(m-2^{n+1})(m+2^{n+1})$$
So, there is a non-negative integer $a$ such that
$$m+2^{n+1}=p^{k-a},m-2^{n+1}=p^a \Longrightarrow 2^{n+2}=p^a(p^{k-2a}-1)$$
Since $\gcd(2,p)=1$, we get $a=0$ to have
$$ p^{k}-2^{n+2}=1$$
If $2^{n+2}+1$ is prime, then $k=1$.
If $2^{n+2}+1$ is not prime, then $k\ge 2$. By Catalan's conjecture (or Mihăilescu's theorem), there is no solution.
Remark 2: Note that when $z = 2^{2n}$, there is a solution only if the Descartes-Frenicle-Sorli Conjecture that $k=1$ on odd perfect numbers holds. But even then, note that
$$m = 2^{n+1}+1 < 2^{n+2}+1 = p$$
which would contradict the Dris conjecture.
Final Notes:
From the following MSE question:
it appears that MSE user FredH has already been able to come up with an unconditional proof for the problem as stated in the title.
We reproduce FredH's proof here:
Here's a way to finish the proof without appealing to any conjecture.
If $q^k n^2$ is a perfect number with $\gcd(q,n)=1$, 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
$$2(q^{k+1}-1) \sigma(n^2) = (q-1) q^k (q^k + 1)^2. (*)$$
Consider the GCD of $q^{k+1}-1$ with the right-hand side:
$$\gcd(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)\gcd(q^{k+1}-1,q^k+1)^2,$$
since $q^k$ is coprime to $q^{k+1} - 1$.
Noticing that $q^{k+1} - 1$ = $q(q^k + 1) - (q + 1)$, we find $\gcd(q^{k+1}-1,q^k+1) = \gcd(q+1,q^k+1)$, which is $q+1$ because $k$ is odd.
Thus
$$\gcd(q^{k+1}-1, (q-1) q^k (q^k + 1)^2) \le (q-1)(q+1)^2.$$
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.
