If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

(Note:  This post was copied verbatim from this MSE question.)

Let $\sigma(x)$ be the sum of the divisors of $x$.  Denote the deficiency of $x$ by $D(x) := 2x - \sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x) := \sigma(x) - x$.

Here is my question:

If $q^k n^2$ is an odd perfect number with Euler prime $q$, does this equation imply that $k=1$?

$$D(q^k)D(n^2)=2s(q^k)s(n^2)$$

I only know that $k=1$ is true if and only if (one of) the following conditions hold:

(1) $\sigma(n^2)/q \mid n^2$

(2) $D(n^2) \mid n^2$

(3) $\gcd(n^2, \sigma(n^2)) = D(n^2)$

Source of Equation

From the fundamental equation

$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$

we obtain

$$\frac{2n^2 - \sigma(n^2)}{\sigma(q^k) - q^k} = \frac{D(n^2)}{s(q^k)} = \gcd(n^2, \sigma(n^2))$$

and

$$\frac{\sigma(n^2) - n^2}{\frac{2q^k - \sigma(q^k)}{2}} = \frac{2s(n^2)}{D(q^k)} = \gcd(n^2, \sigma(n^2)),$$

using the identity

$$\frac{A}{B}=\frac{C}{D}=\frac{C-A}{D-B}.$$

Reference

Conditions Equivalent to the Descartes-Frenicle-Sorli Conjecture on Odd Perfect Numbers