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 one would need to make essential use of "true" primality (and not merely "spoof" primality) in some way. It seems like all of the GCD manipulation is compatible with spoof factorizations, if we replace GCD with its spoof-like counterpart.)