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?
 A: Hint 1: If $x>1$ then reducing mod $4$ shows that $y$ is even, say $y=2v$, and so
$$2^x=(3^z+7^v)(3^z-7^v).$$
Hint 2: It follows that $3^z-2^{x-2}=1$.
A: It is easy to verify computationally that considering the equation modulo $17043520$ implies that it has the only solutions:
$$(x,y,z)\in \{ (1, 1, 1),\ (3, 0, 1),\ (5, 2, 2)\}.$$
I refer to the discussion at AOPS on how to find such a modulus. The one I mentioned above may be not the smallest possible, but it does the job.
UPDATE. As established by brute-force, the smallest working modulus here is $128320$.
PS. For the theory behind such equations, see the paper:

J. L. Brenner, L. L. Foster "Exponential Diophantine equations".
Pacific J. Math. Volume 101, Number 2 (1982), 263-301.
http://projecteuclid.org/euclid.pjm/1102724775

A: 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 $p > 0$
A: 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 $z=2$.
A: Refer to Henri Cohen's Number Theory Volume I: Tools for Diophantine Equations
p. $410-411$ and also $461$ for exercise $35$.
For $x = 1$, the equation becomes $2+7^{y}=9^{z} \implies 2+7^{y}=(3^z)^2 \implies (3^{z})^{2}=7^{y}+2.$
On page $410$ we have $6.7$ The Equation:
$y^{2} = x^{n}+t$ for $-100 \leq t \leq -1$ and $n$ even, $t \not\equiv 1 \pmod {8}$, and $t$ squarefree.
By page $419$, the above result is generalized to include odd $n$ as well, showing that each equation satisfying the above has a finite number of solutions, listed in a table in the book.
It is indicated that using the reasoning of the proofs of recent theorems, this result can be extended to $|t| \leq 100.$
It seems a lot of directly related material is in Chapter $6.7$ of this book and the in-text references, as well as the prerequisite material contained within, however it is clear much of it is "left to the reader" to organize as a proof for exercise $35$ which I am personally unable to do myself at this time.
