We write the equality as (z-1)(z+1)=2^x*7^y;We know that both of z-1 and z+1 must be even.Then,we need to check 2 situations:z-1=2^a; z+1=2^b*7^z such as a+b=x and z-1=2^a*7^t;z+1=2^b*7^l such as a+b=x and l+t=y.In first situation we get 2^b*7^y-2^a=2;2(2^(a-1)+1)=2^b*7^y.We certainly get b=1 and 2^(a-1)+1=7^y.There is no solution for a<=3 and for a>=4 we get y even by mod 8,however,then 2^(a-1)=0(mod 3). So,one case left.2^b*7^l-2^a*7^t=2;We divide the statement by 2 and get 2^(b-1)*7^l=2^(a-1)*7^t+1.We get a=1 and l=0.The case we check is 7^t+1=2^(b-1) Only solution here is t=1 b=4.If we put all we have done together we get:(x,y,z)=(5,1,15) as only solution.

Rearranging we get $(z-1)(z+1)=2^x7^y$. Since $x\geq 1$, it follows that $z-1$ and $z+1$ are both even. Notice that either $\nu_2(z-1)=1$ or $\nu_2(z+1)=1$, and only one of $z-1$ and $z+1$ can be divisible by $7$. Thus we must have $(z-1,z+1)=(2^{x-1},2\cdot 7^y)$ or $(z-1,z+1)=(2\cdot 7^y,2^{x-1})$, since $z-1=2$ or $z+1=2$ yields no solutions. If $(z-1,z+1)=(2^{x-1},2\cdot 7^y)$, we have $2^{x-2}+1=7^y$, which has no solutions because a power of $2$ is never $-1$ modulo $7$. So assume $(z-1,z+1)=(2\cdot 7^y,2^{x-1})$. Then $2^{x-2}-1=7^y$. Let $p$ be a prime factor of $x-2$ so $2^p-1\mid 2^{x-1}-1$. It follows that $2^p\equiv 1\pmod{7}$ so $3\mid p$. Since $p$ is prime, $p=3$. Thus $x-2$ is a power of $3$. If $y\geq 2$, we have $2^{x-2}\equiv 1\pmod{49}$. Since $\mathrm{ord}_{49}(2)$ divides $\varphi(49)=42$ and is not $3$, it has a prime factor other than $3$. This prime factor divides $x-2$, a contradiction. Thus $y=1$ so $\boxed{(x,y,z)=(5,1,15)}$ is the only solution. $\square$