Find positive integer

We can rewrite the second equation as $xy - 2x - 2y - 2z = 0$. Adding $4$ to both sides, we get $xy - 2x - 2y - 2z + 4 = 4$, which factors as $(x-2)(y-2) - 2(z-2) = 4$. Rearranging, we have $(x-2)(y-2) = 2(z-2) + 4$. Since the left side is even, the right side must be even as well, so $z$ must be even. Let $z = 2k$ for some positive integer $k$. Then, we have $(x-2)(y-2) = 4k - 4$, or $(x-2)(y-2) = 4(k-1)$. Since $k-1$ is positive, we can let $k-1 = a$ for some positive integer $a$, so $(x-2)(y-2) = 4a$. We want to find positive integers $x$ and $y$, so we can let $x-2 = 4$ and $y-2 = a$, or $x = 6$ and $y = a+2$. Since $x^2 + y^2 = z^2$, we have $36 + (a+2)^2 = (2k)^2$, or $a^2 + 4a + 4 + 36 = 4k^2$, or $a^2 + 4a + 40 = 4k^2$. Rearranging, we get $a^2 + 4a + 4 - 4k^2 + 36 = 0$, or $(a+2)^2 - 4k^2 + 36 = 0$. This factors as $(a+2)^2 - (2k)^2 = (a+2-2k)(a+2+2k) = 0$. Since $a+2-2k$ and $a+2+2k$ are both positive, we must have $a+2-2k = 1$ and $a+2+2k = a+4 = 2k$, or $a = 2k-4$. Since $a = k-1$, we have $k-1 = 2k-4$, or $k = 3$. Substituting back in, we have $z = 2k = 6$, $x = 6$, and $y = a+2 = 2k-4+2 = 2k-2 = 4$. Thus, the positive integers $x$, $y$, and $z$ that satisfy the given conditions are $\boxed{6, 4, 6}$.