$3^n  2^m = \pm 41$ is not possible for integers $n$ and $m$. How to prove it?

The congruence $3^n  2^m \equiv 41\pmod{60}$ has no solutions. The congruence $3^n  2^m \equiv 41\pmod{72}$ has no solutions. 


As a valuable hint for solving the problem, I consider the following extract from my lectures on elementary number theory. Theorem ($\approx1320$; Levi ben Gerson 12881344). The equations $$ (1) \quad 3^p2^q=1 $$ and $$ (2) \quad 2^p3^q=1 $$ have no solutions in integers $p,q>1$, except the solution $p=2$, $q=3$ to equation (1). Proof. (1) If $p=2k+1$, then $$ 2^q=3^p1=3\cdot9^k1\equiv2\pmod4, $$ which is impossible for $q>1$. If $p=2k$, then $2^q=3^p1=(3^k1)(3^k+1)$ implying $3^k1=2^u$ and $3^k+1=2^v$. Since $2^v2^u=(3^k+1)(3^k1)=2$, we have $v=2$ and $u=1$. This corresponds to the unique solution $q=u+v=3$ and $p=2$. (2) If $q\ge1$, then $3^q+1$ is not divisible by~$8$. Indeed, if $q=2k$, then $3^q+1=9^k+1\equiv2\pmod8$; and if $q=2k+1$, then $3^q+1=3\cdot9^k+1\equiv4\pmod8$. Therefore $p\le2$, hence $p=2$. The latter implies $q=1$ which does not correspond to a solution. 

