ABC Conjecture: $ABC=2^m{p}^n{q}^r$ with Fermat or Mersenne Primes

For $p$ and $q$ any two distinct Fermat or Mersenne primes, $m,n,r$ as positive integers and $\mu =\pm 1$ satisfying any diophantine relation, $(i)2^m+\mu =p^n{q}^r,(ii)2^m{p}^n+\mu =q^{r}or(iii)p^n+\mu q^r=2^m$, it is shown that the number of triplets $\{A,B,C\}$ with $\gcd (A,B)=1$ and $C=A+B$, for which their product is of the form $ABC=2^m{p}^n{q}^r$ and which satisfy $C>\mathrm{r}\mathrm{a}\mathrm{d}(ABC{)}^{1+\epsilon }$ for any real $\epsilon >0$, is finite. For the triplet $\{2^{y+1},2^{2y}+1,(2^y+1{)}^2\}$, a solution to (iii) with positive integer $y$ such that $2^y+1$ and $2^{2y}+1$ are primes, $\mathrm{r}\mathrm{a}\mathrm{d}(ABC{)}^{1+\epsilon }>C$ holds for any $\epsilon >0$. Furthermore, finiteness of the number of solutions of (iii) when $n$ is even, is demonstrated elsewhere (Ref. [64]). All other solutions are enumerated.