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, when $p^k m^2$ is an odd perfect number with special/Euler prime $p$.
To begin with, write
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{\sigma(p^{k-1})}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
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
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)}{p^k - (p^k - 1)},$$
from which we get
$$\gcd(m^2,\sigma(m^2))=\sigma(m^2)-(2m^2 - \sigma(m^2))(p-1)=2m^2 - p(2m^2 - \sigma(m^2))$$
$$= 2m^2 - pD(m^2),$$
or equivalently,
$$\gcd(m^2,\sigma(m^2))=2(1 - p)m^2 + p\sigma(m^2).$$
