If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that $q n^2$ is perfect? Let $q$ be an (odd) prime, and let $\gcd(q,n)=1$.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
A number $N$ is said to be perfect if $\sigma(N)=2N$.
Here is my question:

If $\gcd(n^2, \sigma(n^2)) = q\sigma(n^2) - 2(q - 1) n^2$, does it follow that $q n^2$ is perfect?


CONTEXT
It is known that if $p^k m^2$ is perfect, where $p$ is an odd prime satisfying $\gcd(p,m)=1$, then we obtain
$$i(p):=\dfrac{\sigma(m^2)}{p^k}=\dfrac{2m^2}{\sigma(p^k)}=\dfrac{D(m^2)}{s(p^k)},   \tag{*}$$
where $D(x)=2x-\sigma(x)$ is the deficiency of $x$ and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
(Note that $i(p)$ is odd.)  We then obtain
$$\sigma(m^2) = p^k {i(p)}$$
and
$$m^2 = \dfrac{\sigma(p^k)}{2}\cdot{i(p)},$$
so that
$$\gcd(m^2, \sigma(m^2)) = i(p)\cdot\gcd\Bigg(p^k, \dfrac{\sigma(p^k)}{2}\Bigg) = i(p).$$
Additionally, we can also rewrite Equation (*) as
$$i(p) = \dfrac{\sigma(m^2)}{p^k} = \dfrac{D(m^2)}{s(p^k)} = \dfrac{(p-1)D(m^2)}{p^k - 1} = \dfrac{\sigma(m^2) - (p-1)D(m^2)}{p^k - (p^k - 1)}$$
$$= p\sigma(m^2) - 2(p-1)m^2.$$
Specializing to $k=1$, we obtain the implication
$$qn^2 \text{ is perfect with } \gcd(q,n)=1 \implies \gcd(n^2, \sigma(n^2))=q\sigma(n^2) - 2(q-1)n^2.$$

SANITY CHECK
Let $qn^2$ be an even perfect number.  Then $q = 2^t - 1$ and $n^2 = 2^{t-1}$, for some prime number $t$.
We compute
$$\gcd(n^2, \sigma(n^2)) = \gcd(2^{t-1}, 2^t - 1) = 1$$
and
$$q\sigma(n^2) - 2(q - 1)n^2 = (2^t - 1)(2^t - 1) - 2(2^t - 2){2^{t-1}} = 2^{2t} - 2^{t+1} + 1 - 2^{2t} + 2^{t+1} = 1,$$
whence we have equality between $\gcd(n^2, \sigma(n^2))$ and $q\sigma(n^2) - 2(q - 1)n^2$.
(Note that these computations also "work" for the even perfect number $6$, even though it is squarefree.)

PROBLEM
This takes care of one direction.  Can you come up with a proof for the other direction?
Alas, this is where I get stuck.
 A: Note that, in order to prove that $qn^2$ is perfect (with $\gcd(q,n)=1$) follows from $$\gcd(n^2,\sigma(n^2))=q\sigma(n^2)-2(q-1)n^2,$$
it suffices to show that $I(q)I(n^2)=2$ where $I(x)=\sigma(x)/x$ is the abundancy index of the positive integer $x$.
But we have the biconditional
$$I(q)I(n^2) = 2 \iff \gcd(n^2,\sigma(n^2))=q\sigma(n^2)-2(q-1)n^2=qn^2\Bigg(I(n^2)-\dfrac{2(q-1)}{q}\Bigg) = qn^2 \Bigg(\dfrac{2q}{q+1}-\dfrac{2(q-1)}{q}\Bigg)=qn^2 \Bigg(\dfrac{2}{q(q+1)}\Bigg) = \dfrac{2n^2}{q+1} = \dfrac{\sigma(n^2)}{q} = D(n^2),$$
where the last three ($3$) equations do hold for a perfect number satisfying $I(q)I(n^2)=2$.

HOWEVER, we also have, given an integer $1 \neq k \equiv 1 \pmod 4$ and a prime $p \equiv 1 \pmod 4$ satisfying $\gcd(p,m)=1$, that
$$I(p^k)I(m^2)=2 \iff \gcd(m^2,\sigma(m^2))=p\sigma(m^2)-2(p-1)m^2=pm^2\Bigg(I(m^2)-\dfrac{2(p-1)}{p}\Bigg) = pm^2 \Bigg(\dfrac{2}{I(p^k)}-\dfrac{2(p-1)}{p}\Bigg)=pm^2 \Bigg(\dfrac{2}{p\sigma(p^k)}\Bigg)$$
$$= \dfrac{2m^2}{\sigma(p^k)} = \dfrac{\sigma(m^2)}{p^k} = \dfrac{D(m^2)}{s(p^k)},$$
where the last three (3) equations do hold for a perfect number satisfying $I(p^k)I(m^2)=2$.

Hence, we conclude that the condition
$$\gcd(n^2,\sigma(n^2))=q\sigma(n^2)-2(q-1)n^2  \tag{1}$$
by itself cannot force $qn^2$ is perfect, where $q$ is an odd prime satisfying $\gcd(q,n)=1$, since Equation (1) is also true for perfect numbers $p^k m^2$ with $1 < k \equiv 1 \pmod 4$, $p$ a prime satisfying $p \equiv 1 \pmod 4$, and $\gcd(p,m)=1$, in the sense that while the condition
$$\gcd(m^2,\sigma(m^2))=p\sigma(m^2)-2(p-1)m^2$$
does hold, $pm^2$ is no longer perfect (in fact, it is deficient since it is a proper factor of the perfect number $p^k m^2$, where $k > 1$).
