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

By # 4185 we know that both $x$ and $z$ must be even. Let $x=2k$ and $z=2p$ gives $$\left(3^k\right)^2 + \left(2^y\right)^2 = \left(5^p\right)^2$$

Therefore $(3^k, 2^y, 5^p)$ forms a Pythagorean triplet. This means that there exist distinct integer $m$ and $n$ such that (note that $3^k$ is odd thus cannot be $2mn$) $$\left\{ \begin{array}{lcl} 3^k &=& m^2 - n^2 \\ 2^y &=&2mn \\ 5^p &=& m^2+n^2 \end{array}\right.$$

Because $2^p=2mn$, both $m$ and $n$ must be some powers of $2$. Let $m=2^t$ and $n=2^s$ where integers $t > s$. Setting them back to the previous system yields $$\left\{ \begin{array}{lclcl} 3^k &=& 2^{2t} - 2^{2s} &=&2^{2s}\left(2^{2(t-s)}-1\right) \\ 5^p &=& 2^{2t}+2^{2s} &=&2^{2s}\left(2^{2(t-s)}+1\right) \end{array}\right.$$

Since both $3^k$ and $5^p$ are odd, $s$ must be $0$ which will lead to $t=1$. These means $(m, n)=(2, 1)$ or  $x=y=z=2$.