= Solution
Put $\Lambda=n\log2-m\log3$. It is nonzero by <unique prime factorization>. If $3^m\leq2^{n-1}$ or $3^m\geq2^{n+1}$, the claimed inequality follows immediately after increasing the effective constant. We may therefore assume
$$
2^{n-1}<3^m<2^{n+1},
$$
which implies $m\asymp n$.
The <Baker lower bound for a homogeneous linear form in logarithms>, with the fixed algebraic numbers $2$ and $3$, gives
$$
|\Lambda|\geq n^{-C_0}
$$
for an effective absolute constant $C_0$. If $|\Lambda|>1$, the desired conclusion is again immediate. Otherwise, the <mean value theorem> applied to the <exponential function> on $[-1,1]$ gives $|e^u-1|\geq c|u|$. Hence
$$
|2^n-3^m|
=3^m|e^\Lambda-1|
\geq c3^m|\Lambda|
\geq c2^{n-1}n^{-C_0}.
$$
Absorbing the fixed factor into a larger exponent proves $|2^n-3^m|\geq2^n/n^C$.
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