When $m=1$, $2^{3^{m}}+1=9=3^{2}$, so $k \leqslant 2$. Also, since
$$
2^{3^{m}}+1=\left(2^{3}\right)^{3^{m-1}}+1 \equiv(-1)^{3^{m-1}}+1 \equiv 0\left(\bmod 3^{2}\right) \text {, }
$$

for any positive integer $m, 3^{2} \mid\left(2^{3^{m}}+1\right)$, hence the value of $k$ is 2.
Example 5 (1992 "Friendship Cup" International Mathematics Competition) Find the maximum positive integer $k$, such that $3^{k} \mid\left(2^{3^{m}}+1\right)$, where $m$ is any positive integer.