Example 16 Find all positive integer triples $(x, y, z)$ that satisfy the equation
$$
8^{x}+15^{y}=17^{z}
$$
Taking $\bmod 8$ on both sides of (1) yields $(-1)^{y} \equiv 1(\bmod 8)$, so $y$ is even. Taking $\bmod 7$ yields $2 \equiv 3^{z}(\bmod 7)$, so $z$ is also even. 
At this point, let $y=2 m, z=2 t(m, t \in \mathbf{N})$.

Thus, from (1) we have: $2^{3 x}=\left(17^{t}-15^{m}\right)\left(17^{t}+15^{m}\right)$.
By the fundamental theorem of arithmetic: $17^{t}-15^{m}=2^{x}, 17^{t}+15^{m}=2^{3 x-s}$, hence
$17^{t}=\frac{1}{2}\left(2^{s}+2^{3 x-s}\right)=2^{s-1}+2^{3 x-s-1}$.
Noting that 17 is odd, for (2) to hold, it must be that $s=1$, thus
$17^{t}-15^{m}=2$.
When $m \geqslant 2$, taking $\bmod 9$ on both sides of (3) yields $(-1)^{t} \equiv 2(\bmod 9)$, which is clearly not true, so $m=1$, hence $t=1, x=2$. Therefore, the equation (1) has only one set of positive integer solutions $(x, y, z)=(2,2,2)$.

Note: When dealing with such "exponential" indeterminate equations, a common method is to take a modulus. This allows the use of congruence properties to derive the properties (such as parity) of the solutions or related quantities, simplifying the problem. The key to this method is to choose a good modulus so that taking the modulus truly yields some relationships. This requires that the congruence operations due to taking the modulus are simple. In this problem, the first modulus taken is $\bmod 8$, because $15 \equiv-1(\bmod 8), 17 \equiv 1(\bmod 8)$.