12.16. Solve the equation $2^{n} + 1 = 3^{m}$ in natural numbers.
12.16. Answer: $n=1, m=1$ or $n=3, m=2$.

If $n=1$, then $m=1$. Now consider the case when $n>1$. In this case $3^{m} \equiv 1(\bmod 4)$. Note that $3^{2 k+1}=3 \cdot 9^{k} \equiv 3(\bmod 4)$. Therefore, $m=2 k$, which means $\left(3^{k}-1\right)\left(3^{k}+1\right)=2^{n}$. Thus, the numbers $3^{k}-1$ and $3^{k}+1$ are powers of two, differing from each other by 2. This is only possible when $k=1$, i.e., $m=2$ and $n=3$.