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?

If we allow $y$ and $z$ to be zero, this can be rephrased as: 

For which positive integers $x$ is $2^{x} - 1$ is not divisible by any prime other than $3$ or $7$?

Call the set of such integers $S$.
Suppose $x \in S$. If $t$ divides $x$, then $2^{t} - 1$ divides $2^{x} - 1$, so $t \in S$ as well.
$3$ divides $2^{x} - 1$ iff $x$ is even, and $7$ divides $2^{x} - 1$ iff $x$ is divisible by 3. So we see that if $x \in S$, we must have either $x$ even, $x$ divisible by 3, or $2^{x} - 1 = 1$, i.e. $x = 1$. 
Then if $x \in S$ has a prime factor $p$, $p \in S$, and so we must have $p = 2$ or $p = 3$.
Finally, we see $2^{4} - 1 = 15$ and $2^{9} - 1 = 511$, hence $4$ and $9$ are not in $S$, so no element of $S$ is divisible by $4$ or $9$. The only remaining integers are $1,2,3,6$. We can check $2^{1} - 1 = 1$, $2^{2} - 1 = 3$, $2^{3} - 1 = 7$, and $2^{6} - 1 = 63 = 3^{2} \cdot 7$. So $S = \{1,2,3,6\}$.
Finally, $x = 6$ is the only one which gives $y,z$ positive.
