Is it possible to simplify this expression even further? (Preamble:  This question is tangentially related to this earlier one.)
Let $\sigma(z)$ denote the sum of the divisors of $z \in \mathbb{N}$, the set of positive integers.  Denote the deficiency of $z$ by $D(z):=2z-\sigma(z)$, and the sum of the aliquot divisors of $z$ by $s(z):=\sigma(z)-z$.  Finally, let the abundancy index of $z$ be denoted by $I(z):=\sigma(z)/z$.
If $n$ is odd and $\sigma(n)=2n$, then $n$ is said to be an odd perfect number.  Euler proved that an odd perfect number, if one exists, must have the form $n = p^k m^2$, where $p$ is the special / Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Starting from the fundamental equality
$$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)}$$
one can derive
$$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \gcd(m^2, \sigma(m^2))$$
so that we ultimately have
$$\frac{D(m^2)}{s(p^k)} = \frac{2m^2 - \sigma(m^2)}{\sigma(p^k) - p^k} = \gcd(m^2, \sigma(m^2))$$
and
$$\frac{s(m^2)}{D(p^k)/2} = \frac{\sigma(m^2) - m^2}{p^k - \frac{\sigma(p^k)}{2}} = \gcd(m^2, \sigma(m^2)),$$
whereby we obtain
$$\frac{D(p^k)D(m^2)}{s(p^k)s(m^2)} = 2.$$
We focus on what we can derive from
$$\frac{\sigma(m^2)}{p^k} = \frac{2m^2}{\sigma(p^k)} = \frac{D(m^2)}{s(p^k)} = \gcd(m^2,\sigma(m^2)).$$
We obtain
$$2m^2 - \sigma(m^2) = D(m^2) = s(p^k)\gcd(m^2,\sigma(m^2)) = (\sigma(p^k) - p^k)\gcd(m^2,\sigma(m^2)) = \sigma(p^k)\frac{\sigma(m^2)}{p^k} - {p^k}\frac{2m^2}{\sigma(p^k)} = I(p^k)\sigma(m^2) - \frac{2m^2}{I(p^k)}.$$
Thus, we get
$$\gcd(m^2,\sigma(m^2)) = \frac{D(m^2)}{s(p^k)} = \frac{I(p^k)\sigma(m^2) - \frac{2m^2}{I(p^k)}}{s(p^k)}.$$
We therefore have
$$\gcd(m^2,\sigma(m^2)) = \frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2).$$
Here is my question:

Is it possible to express
  $$\gcd(m^2,\sigma(m^2)) = \frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2)$$
  as an integral linear combination of $m^2$ and $\sigma(m^2)$ (in terms, of course, of $p$ and $k$)?

Sanity Check
When $k=1$, I have
$$\gcd(m^2,\sigma(m^2)) = D(m^2) = 2m^2 - \sigma(m^2),$$
since $s(p^k)=1$ when $k=1$.
When $k=1$, I obtain
$$\frac{I(p^k)}{s(p^k)}\sigma(m^2) - \frac{1}{I(p^k)s(p^k)}(2m^2) = I(p)\sigma(m^2) - \frac{1}{I(p)}(2m^2) = \frac{p+1}{p}\sigma(m^2) - \frac{2p}{p+1}(m^2).$$
Since $p^k m^2 = pm^2$ is assumed to be a(n) (odd) perfect number, then $I(p)I(m^2) = 2$, from which we get
$$I(p) = \frac{2}{I(m^2)}   \text{  and  }  I(m^2) = \frac{2}{I(p)}.$$
Hence,
$$\frac{p+1}{p}\sigma(m^2) - \frac{2p}{p+1}(m^2)$$
simplifies to
$$\frac{2}{I(m^2)}\sigma(m^2) - I(m^2){m^2} = 2m^2 - \sigma(m^2).$$
 A: (This is not a complete answer, just some thoughts that are too long to fit in the comments section.)
So essentially we have to simplify
$$\gcd(m^2,\sigma(m^2))={\frac{I(p^k)}{s(p^k)}}{\sigma(m^2)}-{\frac{1}{I(p^k)s(p^k)}}(2{m^2}).$$
It suffices to consider
$$\frac{I(p^k)}{s(p^k)}$$
and
$$\frac{2}{I(p^k)s(p^k)}$$
separately.
We have
$$\frac{I(p^k)}{s(p^k)} = \frac{\frac{p^{k+1}-1}{{p^k}(p-1)}}{\frac{{p^k}-1}{p-1}}=\frac{p^{k+1}-1}{{p^k}({p^k}-1)}.$$
Notice that
$$\gcd(p^{k+1}-1,p^k)=\gcd\bigg(\frac{p^{k+1}-1}{p-1},\frac{p^k - 1}{p-1}\bigg)=1,$$
so that
$$\frac{I(p^k)}{s(p^k)}$$
is never a positive integer.
Similarly, we know that $I(p^k) > 1$ and $s(p^k) \geq 1$, so that we obtain
$$I(p^k)s(p^k) > 1.$$
This implies that
$$\frac{2}{I(p^k)s(p^k)} < 2,$$
so that if
$$\frac{2}{I(p^k)s(p^k)}$$
were a positive integer, it would be equal to $1$.
Suppose that
$$\frac{2}{I(p^k)s(p^k)}=1.$$
Then we have
$$2 = I(p^k)s(p^k) = \frac{p^{k+1}-1}{{p^k}(p-1)}\cdot{\frac{p^k - 1}{p-1}},$$
so that we obtain
$$2{p^k}(p-1)^2 = (p^{k+1}-1)(p^k - 1)$$
$$2p^{k+2} - 4p^{k+1} + 2{p^k} = p^{2k+1} - p^{k+1} - p^k + 1$$
$$p^{2k+1} - 2p^{k+2} + 3p^{k+1} - 3{p^k} = -1,$$
the last equation of which is a contradiction, as $p^k$ divides the LHS, while $5 \leq p^k$ does not divide the RHS.
We conclude that
$$\frac{2}{I(p^k)s(p^k)}$$
is likewise not a positive integer.

Does this mean that we cannot express
  $$\gcd(m^2,\sigma(m^2))={\frac{I(p^k)}{s(p^k)}}{\sigma(m^2)}-{\frac{1}{I(p^k)s(p^k)}}(2{m^2})$$
  as an integral linear combination of $m^2$ and $\sigma(m^2)$ (in terms of $p$ and $k$)?

A: 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
$$\gcd(m^2,\sigma(m^2))=\frac{\sigma(m^2)}{p^k}=\frac{D(m^2)}{\sigma(p^{k-1})}=\frac{(2m^2 - \sigma(m^2))(p-1)}{p^k - 1}.$$
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,
$$\gcd(m^2,\sigma(m^2))=2(1 - p)m^2  + p\sigma(m^2).$$ 
