Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.
A number $M$ is said to be perfect if $\sigma(M)=2M$. For example, $6$ and $28$ are perfect since $$\sigma(6) = 1 + 2 + 3 + 6 = 2\cdot{6}$$ and $$\sigma(28) = 1 + 2 + 4 + 7 + 14 + 28 = 2\cdot{28}.$$
It is currently unknown if there are infinitely many even perfect numbers. It is also an open problem whether any odd perfect numbers exist. It is widely believed that there are no odd perfect numbers.
Euler proved that an odd perfect number $N$, if one exists, must necessarily have the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
QUESTION
Here is my question:
Can the odd perfect number $N$ be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n}?$$
I would certainly appreciate it if anybody could point me to papers/articles/publications in the literature where this particular inquiry is covered.
CONTEXT
Slowak (1999) proved that the odd perfect number $N$ must be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{d},$$ where $d > 1$.
Dris (2017) showed further that $d$ must have the form $$\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$.
Edit (01/31/2022 - 10:42 PM Manila time) Please check my proof attempt for $\sigma(n^2) \neq q^k n$ below.