There are angles $A, B$, and $C$ in the interval $(0^\circ, 180^\circ)$ such that
$$x = \tan \frac{A}{2}, \quad y = \tan \frac{B}{2}, \quad z = \tan \frac{C}{2}.$$
By the addition and subtraction formulas, the second equation in the given system becomes
$$\begin{aligned} 1 &= \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{B}{2} \tan \frac{C}{2} + \tan \frac{C}{2} \tan \frac{A}{2} \\ &= \tan \frac{A}{2} \tan \frac{B}{2} + \tan \frac{C}{2} \tan \frac{A+B}{2} \left(1 - \tan \frac{A}{2} \tan \frac{B}{2}\right), \end{aligned}$$
$$(1 - \tan \frac{C}{2} \tan \frac{A+B}{2}) (1 - \tan \frac{A}{2} \tan \frac{B}{2}) = 0.$$
Note that $\tan \frac{A}{2} \tan \frac{B}{2} = xy \neq 1$, otherwise $z(x+y) = 0$, implying $z = 0$, which is impossible. Therefore, $\tan \frac{C}{2} \tan \frac{A+B}{2} = 1$, or $\frac{A+B}{2} + \frac{C}{2} = 90^\circ$. In other words, $A, B, C$ are the angles of a triangle. Let $ABC$ denote that triangle.

We rewrite the first equation in the given system as
$$\frac{x}{5(x^2+1)} = \frac{y}{12(y^2+1)} = \frac{z}{13(z^2+1)}.$$
Note that, by the double-angle formula, we have
$$\frac{x}{x^2+1} = \frac{\tan \frac{A}{2}}{\tan^2 \frac{A}{2} + 1} = \frac{\tan \frac{A}{2}}{\sec^2 \frac{A}{2}} = \sin \frac{A}{2} \cos \frac{A}{2} = \frac{\sin A}{2},$$
and analogously for the expressions of $y$ and $z$. We conclude that
$$\frac{\sin A}{5} = \frac{\sin B}{12} = \frac{\sin C}{13}.$$
By the sine rule, the sides of triangle $ABC$ are in ratio $5 : 12 : 13$ with $\sin A = \frac{5}{13}$, $\sin B = \frac{12}{13}$ and $\sin C = 1$. Hence $\frac{x}{x^2+1} = \frac{5}{26}$, or $5x^2 - 26x + 5 = 0$, implying that $x = 5$ or $x = \frac{1}{5}$. Likewise, we have $y = \frac{3}{2}$ or $y = \frac{2}{3}$, and $z = 1$. Substituting $z = 1$ into the second equation in the given system leads to $xy + x + y = 1$, implying that $(x, y, z) = (\frac{1}{5}, \frac{2}{3}, 1)$ is the only solution with $x > 0$. Hence $(\frac{1}{5}, \frac{2}{3}, 1)$ and $(-\frac{1}{5}, -\frac{2}{3}, -1)$ are the solutions of the problem.