Let $x=y=z=2$, we get $m \leqslant \frac{3}{2}$.
Below, we prove that the maximum value of $m$ is $\frac{3}{2}$.
Let $x=\frac{1}{a}-1, y=\frac{1}{b}-1, z=\frac{1}{c}-1$, then we have
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1 \text {. }$$

Rearranging gives $a+b+c=1$.
Thus, the problem is transformed into proving:
$$\begin{array}{c} \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{3}{2} \leqslant \frac{b}{a}+\frac{c}{b}+\frac{a}{c}, \\ \text { i.e., } \frac{3}{2} \leqslant\left(\frac{a}{c}-\frac{a}{b+c}\right)+\left(\frac{b}{a}-\frac{b}{c+a}\right)+\left(\frac{c}{b}-\frac{c}{a+b}\right), \end{array}$$

which is equivalent to $\frac{3}{2} \leqslant \frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)}$.
$$\begin{array}{l} \text { and } \frac{a b}{c(b+c)}+\frac{b c}{a(c+a)}+\frac{c a}{b(a+b)} \\ =\frac{(a b)^{2}}{a b c(b+c)}+\frac{(b c)^{2}}{a b c(c+a)}+\frac{(a)^{2}}{a b c(a+b)} \\ \geqslant \frac{(a b+b c+c a)^{2}}{2 a b c(a+b+c)} \\ =\frac{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)}{2 a b c(a+b+c)} \\ \geqslant \frac{a b c(a+b+c)+2 a b c(a+b+c)}{2 a b c(a+b+c)}=\frac{3}{2} . \end{array}$$

In conclusion, the maximum value of $m$ is $\frac{3}{2}$.