Taking modulo $7$, we find $(-1)^y\equiv 2^z\pmod{7}$. Thus, $3\mid z$ and $y$ is even. Next, taking modulo $8$, we have $x$ is odd. We now take modulo 13, while keeping in mind that $3\mid z$. This yields $7^x\equiv 2^z\in\{-8,-1,1,8\}\pmod{13}$. Since $x$ is odd, the only possibilities are $x\equiv \pm 3\pmod{12}$. Consequently, $3\mid x$. Let $x=3a$, $z=3b$, and factorize to have \[ \left(2^b-7^a\right)\left(2^{2b}+2^b\cdot 7^a + 7^{2a}\right)=13^y. \]Letting $d$ to be the g.c.d. of factors above and seeing that $d\mid 13^y$ and $d\mid 3\cdot 7^{2a}$, we find $d=1$. Thus $2^b-7^a=1$. Consequently, $3\mid b$, set $b=3k$ and have $(2^k-1)(2^{2k}+2^k+1)=7^a$. From here, $2^k-1=1$ must hold. This yields $k=1$, $b=3$, $a=1$. This yields $(x,y,z)=(3,2,9)$ to be the only solution.