As I mentioned in another answer, using GCDs and fractions is usually a bad idea.  Here is how I would interpret all of this material.

Let $N$ be an odd perfect number.  Write its prime factorization as $N=\prod_{i=1}^{k}p_i^{e_i}$.  We may assume $p_1\equiv e_1\equiv 1\pmod{4}$, as usual, and all the other exponents are even.  Fix $m:=\prod_{i=2}^{k}p_i^{e_i/2}$. 

From $\sigma(N)=2N$ and the fact that for any prime $p$ we have $p\nmid \sigma(p^e)$, we see that if we write $\sigma(m^2)=\prod_{i=1}^{k}p_i^{f_i}$, then $f_1=e_1$.

Set $s:=\prod_{i=2}^{k}p_i^{f_i}$, so that $\sigma(m^2)=p_1^{e_1}s$.  Also, fix $t:=m^2/s = \prod_{i=2}^{k}p_i^{e_i-f_i}$.  Notice that $\sigma(p_1^{e_1})=2t$.

Now, you defined the quantity $G$, which is
$$
G:=\gcd(\sigma(p_1^{e_1}),\sigma(m^2))=\gcd(2t,p_1^{e_1}s)=\gcd(s,t)=\prod_{i=2}^{k}p_i^{\min(e_i-f_i,f_i)}.
$$
You also defined the quantity $H$, which is
$$
H:=\gcd(m,\sigma(m^2))=\gcd(m,p_1^{e_1}s)=\gcd(m,s)=\prod_{i=2}^{k}p_i^{\min(e_i/2,e_i-f_i)}.
$$
Finally, you defined the quantity $I$, which is
$$
I:=\gcd(m^2,\sigma(m^2))=\gcd(m^2,p_1^{e_1}s)=\gcd(st,s)=s=\prod_{i=2}^{k}p_i^{f_i}.
$$

Thus, these quantities, and **every other equality** in your post boils down to trivial identities involving minimums in terms of $e_i$ and $f_i$.  They provide no new information.  For example, the quantity $x$ you define is
$$
x:=m/H = \prod_{i=2}^{k}p_i^{\max(0,e_i/2-f_i)}.
$$
Your question about whether we can rule out $x>1$, essentially asks whether we can rule out $f_i<e_i/2$.  (And now, one should ask: Can you rule that out in the spoof odd perfect numbers?  If not, then something deep would have to be present to rule it out in actual odd perfect numbers.)