Is this proof regarding odd perfect numbers valid? (Edited in response to a comment.)
Here are my:
QUESTIONS

(1) Is this proof regarding odd perfect numbers valid, particularly the part where it says
$$\dfrac{2n^2}{D(n^2)} \neq (q + 1)?$$


(2) If the proof is incorrect, how can the argument be mended to produce a valid proof?

Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.  (That is, $\gcd(q,n)=1$ and $q \equiv k \equiv 1 \pmod 4$.)  Let $\sigma(x)$ denote the sum of the divisors of $x \in \mathbb{N}$.
Define
$$D(n^2) := 2n^2 - \sigma(n^2)$$
to be the deficiency of the non-Euler part $n^2$.
Define
$$I(n^2) := \dfrac{\sigma(n^2)}{n^2}$$
to be the abundancy index of $n^2$.
Since $N$ is perfect, then we have
$$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2$$
from which it follows that
$$I(N)=2 \iff I(q^k)I(n^2)=2.$$
But since $q$ is prime, $I(q^k)$ can be rewritten as
$$I(q^k) = \dfrac{q^{k+1} - 1}{q^k (q - 1)}$$
which can be bounded as follows (since $k \equiv 1 \pmod 4$ implies that $k \geq 1$)
$$\frac{q+1}{q} = I(q) \leq I(q^k) < \dfrac{q^{k+1}}{q^k (q - 1)} = \frac{q}{q - 1}.$$
This implies that
$$\dfrac{2(q - 1)}{q} < I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2q}{q + 1}$$
from which it follows that
$$\dfrac{2}{q+1} \leq 2 - I(n^2) = \dfrac{D(n^2)}{n^2} < \dfrac{2}{q},$$
which means that
$$q < \dfrac{2n^2}{D(n^2)} \leq (q + 1).$$
Since $2n^2$ is even and $D(n^2) = 2n^2 - \sigma(n^2)$ is always odd (since $n^2$ is a square), then $$\dfrac{2n^2}{D(n^2)}$$
cannot be an integer.  This implies that
$$\dfrac{2n^2}{D(n^2)} \neq (q + 1)$$
which implies that
$$\dfrac{n^2}{D(n^2)} \neq \dfrac{q+1}{2}$$
and
$$2 - I(n^2) = \dfrac{D(n^2)}{n^2} \neq \dfrac{2}{q+1}.$$
Finally, we obtain
$$I(n^2) \neq \bigg(2 - \dfrac{2}{q+1}\bigg) = \dfrac{2q}{q+1}$$
which implies that
$$I(q^k) = \dfrac{2}{I(n^2)} \neq \dfrac{q+1}{q},$$
thus resulting in
$$k \neq 1.$$

(1)
Your proof looks invalid to me.
You are saying "Since $2n^2$ is even and $D(n^2) = 2n^2 - \sigma(n^2)$ is always odd (since $n^2$ is a square), then $\dfrac{2n^2}{D(n^2)}$ cannot be an integer".
This is not true since $\frac{\text{even}}{\text{odd}}$ can be an integer. For example, $\frac{6}{3}=2$.

Since $D(n^2)$ is odd, we have
$$\frac{2n^2}{D(n^2)}\in\mathbb Z\iff D(n^2)\mid n^2$$
