Exponential diophantine equation $2^x+7^y=9^z$. The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions:
\begin{eqnarray*}
(x_1,y_1,z_1)&=&(3,0,1),\\
(x_2,y_2,z_2)&=&(1,1,1)\\
(x_3,y_3,z_3)&=&(5,2,2).\\
\end{eqnarray*}
There are no other natural solutions for $z≤10000$.
It seems that the equation $2^{x}+7^{y}=9^{z}$ has no other solutions in natural numbers. How can this be proven?

This is a PARTIAL answer. What remains to show is that there are not an infinite number of solutions for $x=1$. Assume $x\ge 3$. Then as in Conner's answer,
$$2^x = (3^z-7^v)(3^z+7^v).$$
This gives for some positive integer $a$ the equation
$$2^a(3^z-7^v)=(3^z+7^v).$$
Rearranging terms gives
$$(2^a-1)3^z = (2^a+1)7^v,$$
Note that this implies that $a$ must also be at least 3. So this gives $2^a+1=3^z$. As $a$ is at least $3$ [because one can check that there are no solutions to the above equation for $a \in \{1,2\}$], it follows that $2^a+1=3^z$ and thus in particular $3^z$ must be $1$ mod $8$, and so it follows that $z$ must be even, which gives
$$(3^{z/2}-1)(3^{z/2}+1) = 2^a,$$
which implies both $3^{z/2}-1$, $3^{z/2}+1$, must be powers of $2$. As they differ by only $2$, this is possible only if $z=2$ and $a=3$. So if $x$ is at least $3$, then $v$ must be $1$ and thus $y$ must be $2$. Thus $x$ must be $5$ and $z=2$.
