Case 1: $y\geq 1$ Then $2^x\equiv z^3\pmod 7$.We have that $ord_7(2)=3$ so $2^x\equiv 2,4,1\pmod 7$ and $z^3\equiv 0,1,6\pmod 7$. So we must have $3\mid x$.Let $x=3a$ $9\cdot 7^y=z^3-2^{3a}=(z-2^a)(z^2+z\cdot 2^a+2^{2a})$. If $7\mid gcd (z-2^a,z^2+z\cdot 2^a+2^{2a}$ then $7\mid 3\cdot z\cdot 2^a\Leftrightarrow 7\mid z$,contradiction! So we must have $7\not | z-2^a$(if $7\mid z-2^a $ then $z-2^a\geq 7^y>z^2+z2^a+2^{2a}$,contradiction). This implies that $z-2^a|9$. If $z-2^a=1,z^2+2^az+2^{2a}=9\cdot 7^y$ then $9\cdot7^y=(z-2^a)^2+3z2^a=1+3z2^a\Rightarrow 3|1+3z2^a$,contradiction! If $z-2^a=3,z^2+2^az+2^{2a}=3\cdot7^y$ then $3\cdot 7^y=(z-2^a)^2+3z2^a=9+3z2^a\Leftrightarrow 7^y=3+2^a(2^a+3)$. If $a\geq 3$ then $3\equiv 7^y\equiv \pm 1\pmod8$ contradiction.So $a=0,1,2$ and the only solution is $(x,y,z)=(0,1,4)$ If $z-2^a=9,z^2+2^az+2^{2a}=7^y$ then $3\cdot 7^y=(z-2^a)^2+3z2^a\equiv0\pmod3$,contradiction. So in this case the only solution is $(x,y,z)=(0,1,4)$. Case 2: $y=0$ $2^x+9=z^3\Rightarrow 2^x+2\equiv z^3\equiv 0,\pm 1\pmod 7\Rightarrow 3|x$.Let $x=3a$. $9=(z-2^a)(2^{2a}+z^2+z2^a)$.Since $2^{2a}+z^2+z2^a>z-2^a$ we must have $z-2^a=1,2^{2a}+z^2+z2^a=9\Leftrightarrow 1+32^z=9\Rightarrow 3|1$contradiction So in this case the only solution is $(x,y,z)=(0,1,4)$.