0, 1, 2, or 3

The second equation can be written as $y(x + z) = 37$ . Now as $37$ is prime and $x + z \ge 2$ we must have $y = 1$ , and thus $x + z = 37 \Longrightarrow z = 37 - x$ . The first equation now becomes

$x + xz = 2135 \Longrightarrow x + x(37 - x) = 2135 \Longrightarrow x^{2} - 38x + 2135 = 0$ ,

which has no real solutions, (and thus no positive integer solutions), since the discriminant $38^{2} - 4*2135 \lt 0$ . Thus there are $\boxed{0}$ positive integer solutions $(x,y,z)$ .