Suppose we have a Diophantine equation called (A) and we want to show that it has a finite number of solutions in integers, and suppose we made a manipulation on (A) so it becomes (B).
I ask which one of the following manipulations are allowed i.e if we prove that (B) has a finite number of solutions then (A) has a finite number of solutions .
multiplying or dividing both sides of (A) by a variable which depends on the equation (ex (A) is $n!+1=m^2$ and we divide it by $2^m$)
adding another variable to both side of (A) where this variable doesn't depend on the (A).
We cannot add another variable. Take for example the Diophantine equation $2^n-64=0$. It has finitely many solutions, namely only one solution $n=6$. Adding another variable to both sides changes it to $2^m+2^n-64=2^m$ which has infinitely many solutions $(m,6)$ for all integers $m$.
Conversely however, if the equation $B$ has only finitely many solutions, then cancelling the terms on both sides, also $A$ has only finitely many solutions, because the equations are equivalent.
If we believe in the weak form of Szpiro’s conjecture on elliptic curves, then $$ n!+1=m^2 $$ has indeed only finitely many integer solutions, which probably are $$ (n,m)=(4,5),(5,11),(7,71). $$