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?
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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 $3^z-2^{x-2}=1$.
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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 $z=2$.
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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.
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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 $p > 0$
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: