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?

here is a method I learned from Exponential Diophantine equation $7^y + 2 = 3^x$  and polished. I have answered many questions of the form $A^u - B^v = C$  for given positive integers $A,B,C$
This answer finishes $x=1.$
The following shows that, with $x = 1,$  the largest solution to $9^z - 7^y = 2 $   is $z=y=1.$  First, write $9^z - 9 = 7^y -7$  Next introduce non-negative integers $p,q$  with
$$ 9(9^p - 1) = 7 (7^q - 1), $$
ASSUME $p,q \geq 1$  and reach a contradiction.   We alternate steps, each is a calculation.
$$ 7 | 9^p -1  \Longrightarrow   \; \;  3 | p$$
$$ 9^3 - 1 = 2^3 \cdot 7 \cdot 13 $$
$$ 13 | 7^q -1  \Longrightarrow   \; \;  12 | q$$
$$ 7^{12} - 1 = 2^5 \cdot 3^2 \cdot 5^2   \cdot 13 \cdot 19 \cdot 43 \cdot 181 $$
$$ 19 | 9^p -1  \Longrightarrow   \; \;  9 | p$$
$$ 9^9 - 1 = 2^3 \cdot 7 \cdot 13 \cdot  19 \cdot 37 \cdot 757 $$
$$ 37 | 7^q -1  \Longrightarrow   \; \;  9 | q$$
$$ 7^{9} - 1 = 2 \cdot 3^3  \cdot 19 \cdot 37 \cdot 1063 $$
This number divides $9 (9^p - 1). $  In particular, $27 |  9 (9^p - 1).$
This CONTRADICTS the assumption that $p > 0$
