<s>[INST] Find the integer solutions to $4^x - 9^y = 55$ I want to find the integer solutions of:
$$ 4^x - 9^y = 55$$
For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?
I tried to solve for $y$, but to no avail. WolframAlpha tells me that it is the only solution, but it doesn't provide an explanation.
 [/INST] I like the explanations given in @MishaLavrov's answer, and I understand this only as a comment to his answer.
There, the second sentence introduces a bit obfuscation, I think: the bare idea of needing a crosscheck of $\{0,1,2,3,4,5\} \times 
\{0,1,2,3\}$ insertions is somehow surprising (and misleading) after the introductory statement.
$$4^a - 9^b = 55 $$
This can be factored, and the factors of the lhs can be crosschecked with the two factorizings of the rhs, where the obvious sizes of the factors in the lhs are brought in agreement with the order of sizes of the factors of the rhs:
$$
  (2^a+3^b)(2^a-3^b)= \begin{cases}  (11) \cdot (5) & (a)\\  (55) \cdot (1) &(b) \end{cases} $$
Test $(a)$:
$$ (2^a+3^b)-(2^a-3^b)=2\cdot 3^b \overset?= 11-5 = 6 = 2\cdot 3^1 \implies b=1\\
   (2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 11+5 = 16 = 2 \cdot 2^3 \implies a=3$$
$\qquad \qquad$ both ways of summing the corresponding factors give valide solutions.
Test $(b)$:
$$    (2^a+3^b)-(2^a-3^b)=2\cdot 3^b\overset?= 55-1 = 54 = 2\cdot 3^3 \implies b=3\\
   (2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 55+1 = 56 = 2 \cdot 2^2 \cdot 7 \implies \text{error}
$$
$\qquad \qquad$ the second way of summing the corresponding factors does not agree with the problem.
Conclusion: We have only one solution: $4^3 - 9^1 = 55$
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