In what follows, denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then $n$ is called an odd perfect number.
Euler showed that a hypothetical odd perfect number $n$, if one exists, must have the so-called Eulerian form $$n = p^k m^2$$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Since $\gcd(p,m)=1$ and the divisor sum function $\sigma()$ is multiplicative, then we have $$\sigma(p^k)\sigma(m^2)=\sigma(p^k m^2)=\sigma(n)=2n=2p^k m^2.$$
We ultimately get $$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{m^2}{\sigma(p^k)/2}=\frac{D(m^2)}{s(p^k)}=\frac{2s(m^2)}{D(p^k)} \tag{*},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$ and $s(x)=\sigma(x)-x$ is the aliquot sum of $x$.
We also know that $$\gcd(m^2,\sigma(m^2))=p\sigma(m^2)-2(p-1)m^2.$$
The question is as is in the title:
If $p^k m^2$ is an odd perfect number, then $D(p^k)/s(p^k)$ is in lowest terms. Does this contradict $D(p^k)D(m^2)=2s(p^k)s(m^2)$?
MY ATTEMPT
From this MSE question, we know that $$\gcd(D(p^k),s(p^k)) = 1$$ which means that the fraction $D(p^k)/s(p^k)$ is in lowest terms.
However, from Equation $(*)$, we obtain $$D(p^k)D(m^2)=2s(p^k)s(m^2)$$ which means that the fraction $$\frac{D(p^k)}{s(p^k)}=\frac{2s(m^2)}{D(m^2)}$$ must also be in lowest terms.
This implies that $$\gcd(2s(m^2),D(m^2))=1$$ which is equivalent to the equation $$1=\gcd(2s(m^2),D(m^2))=\gcd(2s(m^2)+D(m^2),D(m^2))$$ $$=\gcd(s(m^2)+s(m^2)+D(m^2),D(m^2))$$ $$=\gcd(s(m^2)+m^2,D(m^2))=\gcd(\sigma(m^2),D(m^2)).$$
But $$1=\gcd(\sigma(m^2),D(m^2))$$ is equivalent to the equation $$1=\gcd(\sigma(m^2),\sigma(m^2)+D(m^2))=\gcd(\sigma(m^2),2m^2)=\gcd(\sigma(m^2),m^2),$$ since $\sigma(m^2)$ is odd.
But the equation $$1=\gcd(\sigma(m^2),m^2)=\frac{\sigma(m^2)}{p^k}$$ contradicts the following result of [Dris, JIS (2012)]: $\sigma(m^2)/p^k \geq 3$.
CONCLUSION
This appears to prove that there are no other perfect numbers, apart from those for which $\gcd(m^2,\sigma(m^2))=1$. Notice that even perfect numbers $(2^q - 1)\cdot{2^{q-1}}$ (except for $6$) are of the form $$P^K M^2$$ where $P=2^q - 1$, $K=1$, and $M = 2^{(q-1)/2}$. Thus, for even perfect numbers, we have $$\gcd(M^2,\sigma(M^2))=\gcd(2^{q-1},\sigma(2^{q-1}))=\gcd(2^{q-1},2^q - 1) = 1.$$
Why? How do you exclude possibilities like $\dfrac 85=\dfrac{24}{15}$ ?
If we have $\gcd(D(p^k),s(p^k))=1$ and $D(p^k)D(m^2)=2s(p^k)s(m^2)$, then we can say that there is an integer $z$ such that $D(m^2)=zs(p^k)$ and $2s(m^2)=zD(p^k)$, which implies that $\gcd(2s(m^2),D(m^2))=z$. However, I don't think that $z=1$ follows from these only.