Since all three equations are cubic, the maximum finite number of solutions is 27. Note that $\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$. Therefore, $z=i\tan\theta\implies y=i\tan 3\theta\implies x=i\tan 9\theta\implies z=i\tan 27\theta$. So $27\theta=\pi k+\theta\implies\theta=\frac{\pi k}{26}$. Therefore we have 24 complex solutions $\left(i\tan\frac{9\pi k}{26},i\tan\frac{3\pi k}{26},i\tan\frac{\pi k}{26}\right)$ for $1\leq k\leq 25$ and $k\neq 13$. In addition, we have 3 trivial real solutions $\boxed{(1,1,1), (-1,-1,-1), (0,0,0)}$.

This is $x=f_y, \; y=f_z, \; z=f_x$ with strictly increasing $f_x=\frac {x(3+x^2)}{1+3x^2}\Rightarrow~x=f^3_x=f_x\Rightarrow~$ $x=y=z\in\{0,1,-1\}$