In general, for $q$ prime and $\gcd(q, n) = 1$, can $N = qn^2$ be almost perfect?

An almost perfect number $N$ is a positive integer satisfying $\sigma(N) = 2N - 1$, where $\sigma(x)$ is the sum of all the divisors of $x$. For example, $\sigma(2) = 1 + 2 = 2\cdot2 - 1$, and thus $2$ is almost perfect.

A number $M$ is perfect if $\sigma(M) = 2M$. Recall that the Eulerian form of an odd perfect number is $M={p^k}{m^2}$. This means the following:

If $k = 1$, then $M=p{m^2}$ is perfect.

If $k > 1$, then $M=p{m^2}$ is deficient.

I was thinking of applying the criterion from this paper1, but alas this is where I get stuck.

To recap, my main (and more specific) question for this post would be:

If $k > 1$, can the divisor $p{m^2}$ of the odd perfect number $M={p^k}{m^2}$ be almost perfect?

I already know that $p^k$, $m$, $pm$ and $m^2$ are not almost perfect if $M={p^k}{m^2}$ is perfect (see paper2).