nt.number theory - Does 2^m = 3^n + r have finitely many solutions for every r?

Yes, it is true that this kind of equation ax+by=c, where a,b,c are non-zero and fixed and x,y are allowed to only have prime factors in a finite set, has only finitely many solutions. This is a special case of Siegel's theorem on integral points on curves.

Your second question may be unknown in the generality you pose. It is interesting that it holds. A remark: if there is a solution to $2^m = 3^n + r$, then $2^{m+kphi(M)} equiv 3^{n+kphi(M)} + r (mod M)$ for all $k,M$ if $(M,6)=1$, so if $M$ exists in this case, then $(M,6)>1$. If there is no solution to the equation $2^m = 3^n + r$, then the existence of $M$ (with $N=0$) is a special case of a conjecture of Skolem.

T. Skolem: Anwendung exponentieller Kongruenzen zum Beweis der
Unlösbarkeit gewisser diophantischer Gleichungen., Avh. Norske Vid. Akad.
Oslo, 12 (1937), 1–16.

Another comment. There are no solutions when $r=11$ but $M=8$ doesn't work since $2^2 equiv 3^2 + 11 mod 8$. $M(11)=205$. (Edit: $M(11)=8$ is OK. I misunderstood the definition, see comments)