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 $M={p^k}{m^2}$, where $p$ is prime with $\gcd(p,m)=1$.  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](http://arxiv.org/abs/1308.6767), 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](http://arxiv.org/abs/1309.0906v9)).