Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
A number $N$ is said to be perfect if $\sigma(N)=2N$. (Equivalently, $N$ is perfect if $I(N)=2$.)
It is known that if $q^k n^2$ is perfect, where $q$ is an odd prime satisfying $\gcd(q,n)=1$, then we obtain $$i(q) := \dfrac{\sigma(n^2)}{q^k} = \dfrac{2n^2}{\sigma(q^k)} = \dfrac{D(n^2)}{s(q^k)} = \dfrac{2s(n^2)}{D(q^k)}, \tag{1}$$ 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 $i(q)$ is odd.) We then obtain $$\sigma(n^2) = q^k \cdot i(q)$$ and $$n^2 = \dfrac{\sigma(q^k)}{2} \cdot i(q),$$ so that $$\gcd(n^2, \sigma(n^2)) = i(q)\cdot\gcd\Bigg(\dfrac{\sigma(q^k)}{2},q^k \Bigg) = i(q). \tag{2}$$
Additionally, we can rewrite Equation (1) as $$i(q) = \dfrac{\sigma(n^2)}{q^k} = \dfrac{D(n^2)}{s(q^k)} = \dfrac{(q - 1)D(n^2)}{q^k - 1} = \dfrac{\sigma(n^2) - (q - 1)D(n^2)}{q^k - (q^k - 1)} = q\sigma(n^2) - 2(q - 1)n^2. \tag{3}$$
Let us consider the equation $$\dfrac{2n^2}{\sigma(q^k)} = q\sigma(n^2) - 2(q - 1)n^2.$$
Dividing through by $qn^2$, we get $$\dfrac{2}{q\sigma(q^k)} = I(n^2) - \dfrac{2(q - 1)}{q},$$ which implies that the reciprocal of $I(n^2) - \dfrac{2(q - 1)}{q}$ is an integer as $$\dfrac{q \sigma(q^k)}{2} = \dfrac{1}{I(n^2) - \dfrac{2(q - 1)}{q}}$$ is an integer (since $\sigma(q^k) \equiv 2 \pmod 4$).
Let us now consider the equation $$\dfrac{D(m^2)}{s(p^k)} = p\sigma(m^2) - 2(p - 1)m^2,$$ for the case of odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $\gcd(p,m)=1$.
Dividing through by $pm^2$, we obtain $$\dfrac{D(m^2)}{pm^2 s(p^k)} = I(m^2) - \dfrac{2(p - 1)}{p}.$$ Taking the reciprocals of both sides, we get $$\dfrac{pm^2 s(p^k)}{D(m^2)} = \dfrac{1}{I(m^2) - \dfrac{2(p - 1)}{p}}, \tag{4}$$ which must be an integer by the considerations in the previous section.
Dividing both sides of Equation (4) by $p$, we obtain $$\dfrac{m^2 s(p^k)}{D(m^2)} = \dfrac{\sigma(p^k)}{2} = \dfrac{1}{p}\cdot\Bigg(\dfrac{1}{I(m^2) - \dfrac{2(p - 1)}{p}}\Bigg), \tag{5}$$ which is still an integer.
Here is my question:
Does it follow from Equation (5) that the divisibility constraint $$D(m^2) \mid m^2$$ must hold?
Note that, for odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $\gcd(p,m)=1$, we actually have the lower bound $$i(p) \geq 3,$$ a result of Dris (2012).
Lastly, note that for even perfect numbers $q n^2$ with Mersenne prime $q$, we actually have $$i(q) = 1.$$