Hereinafter, call a number $N$ perfect if $N$ satisfies $\sigma(N)=2N$, where $$\sigma(x)=\sum_{d \mid x}{d}$$ is the sum of divisors of the positive integer $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$, the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of aliquot divisors of $x$ by $s(x)=\sigma(x)-x$.
Let $n = p^k m^2$ be an odd perfect number given in Eulerian form, that is, $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
We shall use the following results in deriving bounds for $D(m^2)$:
It turns out that it is possible to express $\gcd(m^2,\sigma(m^2))$ as an integral linear combination of $m^2$ and $\sigma(m^2)$, in terms of $p$ alone.
To begin with, write $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$ Now, using the identity $$\frac{A}{B}=\frac{C}{D}=\frac{A-C}{B-D},$$ where $B \neq 0$, $D \neq 0$, and $B \neq D$, we obtain $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2) - (2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)}$$ so that we get $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$
At once, since $s(p^k) \geq 1$, we have the upper bound: $$D(m^2) \leq 2m^2 - pD(m^2) \implies D(m^2) \leq \frac{m^2}{(p+1)/2}.$$ Equality holds if and only if the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
We now attempt to derive a lower bound for $D(m^2)$ (in terms of $p$, $m^2$ and $\sigma(m^2)$), using the result discussed in this recent MSE question:
From the result $$s(a)s(b) + (a + b) \leq s(ab)$$ which holds when $\gcd(a,b)=1$, $a>1$, and $b>1$, then setting $a=p^k$ and $b=m^2$, we obtain $$s(p^k)s(m^2) + (p^k + m^2) \leq s(p^k m^2) = p^k m^2$$ $$\implies 1 + s(p^k)s(m^2) \leq (p^k m^2 - (p^k + m^2) + 1) = (p^k - 1)(m^2 - 1) = (p - 1)(m^2 - 1)s(p^k)$$ $$\implies 1 \leq \bigg((p-1)(m^2 - 1) - s(m^2)\bigg)s(p^k)$$ Multiplying both sides by $D(m^2)$ and dividing through by $s(p^k)$, we get $$\frac{D(m^2)}{s(p^k)} \leq D(m^2)\cdot{\bigg((p-1)(m^2 - 1) - s(m^2)\bigg)}.$$ But we know from a previous calculation that $$\frac{D(m^2)}{s(p^k)}=2m^2 - pD(m^2)=2(1-p)m^2 + p\sigma(m^2).$$ Hence, we have the lower bound $$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2).$$
Summarizing, we have the bounds:
$$\frac{2(1-p)m^2 + p\sigma(m^2)}{(p-1)(m^2 - 1) - (\sigma(m^2) - m^2)} \leq D(m^2) \leq \frac{m^2}{(p+1)/2}.$$
Here are my questions:
(1) Does anybody here have any bright ideas on how to simplify the lower bound for $D(m^2)$?
(2) Are these bounds best-possible?
We can get a better bound.
To get a better bound, we need a better inequality than $\sigma(x)-x\ge 1$.
So, let us find a better inequality on $\sigma(m^2)$.
To find a better lower bound, let us consider $m$ of the form $PQ$ where $P\lt Q$ are distinct primes.
Then, we have $$\begin{align}\sigma(m^2)&\ge (1+P+P^2)(1+Q+Q^2) \\\\&=1+P+P^2+Q+Q^2+PQ(P+Q+1)+P^2Q^2 \\\\&\ge 1+2+2^2+3+3^2+m(2+3+1)+m^2 \\\\&=m^2+6m+19\end{align}$$ from which we have $$m^2-\sigma(m^2)\le -6m-19$$
Using this, we get, similarly as you did, $$\begin{align}&s(p^k)s(m^2) -s(p^km^2)\le -m^2+p^k(-6m-19) \\\\&\implies s(p^k)s(m^2)\le p^km^2-m^2+p^k(-6m-19) \\\\&\implies s(p^k)s(m^2)\le (p^k-1)(m^2-1)+p^k(-6m-18)-1 \\\\&\implies s(p^k)s(m^2)\le (p-1)(m^2-1)s(p^k)+p^k(-6m-18)-1 \\\\&\implies p(6m+18)+1\le ((p-1)(m^2-1)-s(m^2))s(p^k)\end{align}$$ Multiplying the both sides by $\frac{D(m^2)}{s(p^k)}$ gives $$\frac{D(m^2)}{s(p^k)}(p(6m+18)+1)\le ((p-1)(m^2-1)-s(m^2))D(m^2)$$
from which we get $$\frac{(2(1-p)m^2 + p\sigma(m^2))(p(6m+18)+1)}{(p-1)(m^2-1)-(\sigma(m^2)-m^2)}\le D(m^2)$$