Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$ 
Find all positive integers $x,y,z$ that satisfy  $$2^x=3^y7^z+1$$.

I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?

We have to show that for $x\ge 7$, the number $2^x-1$ has a prime divisor different from $3$ and $7$. 
If $x$ has a prime factor $p\ge 5$, then $2^p-1$ divides $2^x-1$ and all prime factors of $2^p-1$ must be of the form $2kp+1$, hence there must be a prime factor greater than $7$. 
Otherwise $x$ must be divisible by $4$ or by $9$ , hence $2^4-1$ or $2^9-1$ must divide $2^x-1$. This implies that $5$ or $73$ must be a prime factor. Since $x<6$ gives no solution, in fact $x=6$ is the only one.
