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.
On
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: $4^3 - 9^1 = 55$
This factors as $(2^x + 3^y)(2^x-3^y) = 55$, so $2^x + 3^y \in \{1, 5, 11, 55\}$. This leaves us only finitely many cases of $x$ and $y$ to check: $x \le 5$ (because $2^6$ is bigger than $55$) and $y \le 3$ (because $3^4$ is bigger than $55$).
A check of all $(x,y) \in \{0,1,2,3,4,5\} \times \{0,1,2,3\}$ gives only one solution: $(x,y) = (3,1)$.
We can be more clever and eliminate some cases, reducing the amount of casework we have to do, but that's already enough.
To minimize casework: first, because $2^x + 3^y > 2^x - 3^y$, we know that $2^x + 3^y$ is either $11$ or $55$, with $2^x - 3^y$ being either $5$ or $1$ respectively. But adding them together gives a value of either $11+5=16$ or $55+1 = 56$ for $2 \cdot 2^x$. Only the first of these is a power of $2$, so we conclude that $2^x + 3^y = 11$.
To get this, $3^y$ can be either $1$, $3$, or $9$. This leaves $10$, $8$, or $2$ for $2^x$. We reject $2^x = 10$ because $x$ must be an integer. We reject $2^x = 2$ with $3^y=9$ because the original equation $4^x-9^y=55$ is not satisfied when $x=1$ and $y=2$. Finally, the case $2^x=8$ with $3^y=3$ gives us the solution $(x,y) = (3,1)$.
It's harder, but still possible, to solve the more general equation $2^x - 3^y = 55$, to which we now know that $(x,y) = (6,2)$ is a solution.
With the aim of rejecting solutions with $y>2$, consider the equation modulo $27$: if $y$ is $3$ more, then $3^y \equiv 0 \pmod{27}$, so $2^x \equiv 55 \equiv 1 \pmod{27}$. The order of $2$ modulo $27$ is $18$, so $2^x \equiv 1 \pmod{27}$ exactly when $x$ is a multiple of $18$.
In that case, take the equation modulo $73$, which is one of the prime factors of $2^{18}-1$, so we know that $2^x \equiv 1 \pmod{73}$. This tells us that $3^y \equiv 1 - 55 \equiv 19 \pmod{73}$. This has no solutions: the powers of $3$ can take on the values $3, 9, 27, 8, 24, 72, 70, 64, 46, 65, 49, 1$, after which they repeat.
So it's impossible to have $y \ge 3$. We can try $y=0$ (in which case $2^x=56$ has no solution), $y=1$ (in which case $2^x = 58$ has no solution), or $y=2$ (in which case $2^x = 64$ gives us $(x,y) = (6,2)$).