Prove that since $m$ is odd, and $1,2, \cdots, m$ is a complete residue system modulo $m$, then $2 \times 1, 2 \times 2, \cdots, 2 \times m$ is also a complete residue system modulo $m$, so,
$$1^{n}+2^{n}+\cdots+m^{n} \equiv (2 \times 1)^{n}+(2 \times 2)^{n}+\cdots+(2 \times m)^{n} \pmod{m}.$$

Thus, $m \mid \left(2^{n}-1\right)\left(1^{n}+2^{n}+\cdots+m^{n}\right)$, and combining with $\left(m, 2^{n}-1\right)=1$, the proposition holds.