Square the two equations to get $(x-yz)^2+(xz+y)^2=x^2-2xyz+y^2z^2+x^2z^2+2xyz+y^2$ $=(x^2+y^2)(z^2+1)=11^2+13^2=290$. The factors of 290 are $1, 2, 5, 10, 29, 58, 145, 290$. We need one of these to be one more than a square: the ones that satisfy this are $1, 2, 5, 10, 145, 290$. We look at these in turn. If $z^2+1=1$, then $z=0$ and $x=11$ and $y=13$. If $z^2+1=2$, then $z=\pm 1$. If $z=1$ then $x=12$ and $y=1$. If $z=-1$ then $x=1$ and $y=12$. If $z^2+1=5$, then $z=\pm 2$. If $z=2$ then $x=7$ and $y=-3$. If $z=-2$ then $x=-\frac{9}{5}$ and $y=\frac{37}{5}$ which aren't integral. If $z^2+1=10$, then $z=\pm 3$. If $z=3$ then $x=\frac{23}{5}$ and $y=-\frac{14}{5}$ which aren't integral. If $z=-3$ then $x=-2$ and $y=5$. If $z^2+1=145$, then $z=\pm 12$. If $z=12$ then $x=1$ and $y=-1$. If $z=-12$ then $x=-\frac{119}{145}$ and $y=\frac{167}{145}$ which aren't integral. If $z^2+1=290$, then $z=\pm 17$. If $z=17$ then $x=\frac{20}{29}$ and $y=-\frac{21}{29}$ which aren't integral. If $z=-17$ then $x=-\frac{3}{5}$ and $y=\frac{4}{5}$ which aren't integral either. So the solutions are $(x, y, z)=(11, 13, 0), (12, 1, 1), (1, 12, -1), (7, -3, 2), (-2, 5, -3), (1, -1, 12)$.