1.31 Find the largest natural number $k$, such that $3^{k}$ divides $2^{3^{m}}+1$, where $m$ is any natural number.
("Friendship Cup" International Mathematical Competition, 1992)
[Solution] When $m=1$, $2^{3^{m}}+1=9$.
From $3^{k} \mid 19$ we know $k \leqslant 2$.
Below we prove that for any positive integer $m$,
$$
\begin{array}{c}
3^{2} \mid 2^{3^{m}}+1 \\
2^{3^{m}}+1 \\
=\left(2^{3}\right)^{3^{m-1}}+1 \\
\equiv(-1)^{3^{m-1}}+1 \\
\equiv 0 \quad(\bmod 9) .
\end{array}
$$

In fact,

So
$$
912^{3^{m}}+1
$$

Thus, the maximum value of $k$ is 2.