Example 2 Let $f, g: \mathbf{Z}^{+} \rightarrow \mathbf{Z}^{+}$ be strictly increasing sequences, and $f\left(\mathbf{Z}^{+}\right) \cap g\left(\mathbf{Z}^{+}\right)=$ $\varnothing, f\left(\mathbf{Z}^{+}\right) \cup g\left(\mathbf{Z}^{+}\right)=\mathbf{Z}^{+}, g(m)=f[f(m)]+1$, find $f(2 m)$.
We take $\alpha=\frac{1+\sqrt{5}}{2}, \beta=\frac{3+\sqrt{5}}{2}$, then it is evident that $\frac{1}{\alpha}+\frac{1}{\beta}=1$, $\beta=\alpha^{2}$, and $\alpha, \beta$ are the unique pair of irrational numbers satisfying the above two conditions. We construct two functions $f(m)=[\alpha m], g(m)=[\beta m]$, which are clearly strictly increasing functions from $\mathbf{Z}^{+}$ to $\mathbf{Z}^{+}$. According to the conclusion of the problem, they also satisfy $f\left(\mathbf{Z}^{+}\right) \cap g\left(\mathbf{Z}^{+}\right)=\varnothing, f\left(\mathbf{Z}^{+}\right) \cup g\left(\mathbf{Z}^{+}\right)=\mathbf{Z}^{+}$. Therefore, we only need to verify that $g(m) = f[f(m)] + 1$. In fact, on one hand,
$$
[\alpha m]1$, so the equality cannot hold);
On the other hand, we have
$$
\begin{array}{c}
\alpha=1+\frac{1}{\alpha}, \beta=1+\alpha, \\
\alpha[\alpha m]=\left(1+\frac{1}{\alpha}\right)[\alpha m]=[\alpha m]+\frac{1}{\alpha}[\alpha m] \\
>[\alpha m]+\frac{1}{\alpha}(\alpha m-1)=[\alpha m]+\frac{1}{\alpha} \alpha m-\frac{1}{\alpha} \\
=[(\alpha+1) m]-\frac{1}{\alpha}>[\beta m]-1,
\end{array}
$$

Thus,
$$
[\alpha(\alpha m)] \geqslant[\beta m]-1 .
$$

Combining the above two aspects, we have $[\alpha(\alpha m)]=[\beta m]-1$, so
$$
\begin{aligned}
f(2 m) & =[\alpha(2 m)]=\left[\frac{1+\sqrt{5}}{2}(2 m)\right] \\
& =[(1+\sqrt{5}) m]=m+[\sqrt{5} m] .
\end{aligned}
$$