Any solution?

Observe that $x,y$ are of different parity, without loss of generality, let $x=2t$ is even and $y$ is odd. We are going to prove that $3|t$. Assume otherwise that $3 \nmid t$. The original equation can be write as \[ y^y=9^z-(2t)^{2t}=\left( 3^z-(2t)^t \right) \left( 3^z+(2t)^t \right). \]If a prime $p$ divides both $3^z-(2t)^t$ and $3^z+(2t)^t$, then $p$ is odd and $p| \gcd \left( 3^z,(2t)^t \right)=1$, contradiction. Hence $\gcd \left( 3^z-(2t)^t,3^z+(2t)^t \right)=1$. So we can write \[ 3^z-(2t)^t=u^y, ~ 3^z+(2t)^t=v^y, \]where $u,v$ are coprime integers and $y=uv$. Adding these two equations, we get \[ 2 \cdot 3^z=u^y+v^y=(u+v) \left( u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1} \right). \]If a prime $q$ divides both $u+v$ and $u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1}$, then \[ 0 \equiv u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1} \equiv yu^{y-1} \equiv u^yv \pmod q. \]So $q|u$ or $q|v$, which implies that $q|\gcd(u,v)$, which is absurd as $u,v$ are coprime. Therefore $\gcd \left( u+v,u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1} \right)=1$, so $\{u+v, u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1} \} = \{2,3^z \} ~\text{or}~ \{1,2 \cdot 3^z \} $. But as $u \ge 1$ and $v \ge 3$ are odd integers, $u+v>2$, and \[ u^{y-1}-u^{y-2}v+\cdots-uv^{y-2}+v^{y-1}=u^{y-1}+(v^{y-2}+\cdots+vu^{y-3})(v-u) \ge 1+3 \cdot 2 >2. \]The final contradiction implies that $3|t$, hence $3|x$ and $3|y$. Therefore, $\left( x^{x/3} \right)^3+\left( y^{y/3} \right)^3=3^{2z}$, which is again impossible due to Zsigmondy theorem.