It is well known that $x=y=z=2$ is one solution. We are going to show that no other solution exists.

Taking MOD $8$ on both sides, $0+(-1)^y\equiv 1\mod{8}$, concludes $y$ must be even.

Taking MOD $7$ on both sides, $1+1\equiv 3^z\mod{7}$, concludes $z$ must be even too.

Let $y=2m$ and $z=2n$, and rearrange the given equation: $$8^x = 17^{2n}-15^{2m}\implies 2^{3x} = (17^n-15^m)(17^n+15^m)$$

As the left side is a power of $2$, the right side must be so as well which means both terms are some powers of $2$. Let's assume $17^n-15^m = 2^t$, then $$\left\{\begin{align}17^n - 15^m &= 2^t\\17^n+15^m&=2^{3x-t}\end{align}\right.$$

Solving this equation system yields $$17^n=(2^{t-1}+2^{3x-t-1}$$

Because $17$ is even, either $2^{t-1}$ or $2^{3x-t-1}$ is odd, i.e. equal $1$.

• If $t-1=0$, the original equation becomes $17^n-15^m=2$. In this case $m-n=1$ is the only solution and yields $x=y=z=2$. (see # 4789.)
• If $3x-t-1=0$, we will have $17^n+15^m=2$ which is impossible.