Example 19 Given $n \geqslant 2, n \in \mathbf{Z}^{+}$, find all $m \in \mathbf{Z}^{+}$, such that for $a_{i} \in \mathbf{R}^{+}, i=1,2$, $\cdots, n$, satisfying $a_{1} a_{2} \cdots a_{n}=1$, then
$$a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} .$$
Let $x=a_{1}=a_{2}=\cdots=a_{n-1}>0, a_{n}=\frac{1}{x^{n-1}}$, then
$$(n-1) x^{m}+\frac{1}{x^{(n-1) m}} \geqslant \frac{n-1}{x}+x^{n-1}$$

From this, we get $m \geqslant n-1$. Now, assume $m \geqslant n-1$, then
$$\begin{aligned}
& (n-1)\left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}\right)+n(m-n+1) \\
= & \left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n-1}^{m}+1+1+\cdots+1\right) (\text{ a total of } m-n+1 \text{ ones}) \\
& +\left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n-2}^{m}+a_{n}^{m}+1+1+\cdots+1\right)+\cdots \\
& \quad+\left(a_{2}^{m}+a_{3}^{m}+\cdots+a_{n}^{m}+1+1+\cdots+1\right) \\
& \geqslant m \sqrt[m]{\left(a_{1} a_{2} \cdots a_{n-1}\right)^{m}}+\cdots+m \sqrt[m]{\left(a_{2} \cdots a_{n}\right)^{m}} \\
& =m\left(a_{1} a_{2} \cdots a_{n-1}+\cdots+a_{2} a_{3} \cdots a_{n}\right)=m\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)
\end{aligned}$$

Therefore,
$$a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{m}{n-1}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)-\frac{n}{n-1}(m-n+1) .$$

Thus, it suffices to prove
$$\begin{array}{c}
\frac{m}{n-1}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)-\frac{n}{n-1}(m-n+1) \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \\
(m-n+1)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n\right) \geqslant 0
\end{array}$$

That is, $\square$
By the assumption and the AM-GM inequality, we get
$$\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n\right) \geqslant n \sqrt[n]{\frac{1}{a_{1}} \cdot \frac{1}{a_{2}} \cdot \cdots \cdot \frac{1}{a_{n}}}-n=0$$

Therefore, the original inequality holds.
Hence, for all $m \geqslant n-1$ and $m \in \mathbf{Z}^{+}$, the inequality is true.