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.)