Find the integer solutions to $4^x - 9^y = 55$
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 $(x,y) = (6,2)$).