Find all pairs of natural numbers $(x, y)$ satisfying:

$$
2^{x}-3^{y}=7
$$
On $2^{x}=7+3^{y} \geqslant 8$ so $x \geqslant 3$. We look modulo 8, we have $3^{y} \equiv 1(\bmod 8)$. But $3^{2 k} \equiv 9^{k} \equiv 1$ $(\bmod 8)$ and $3^{2 k+1} \equiv 3(\bmod 8)$: we deduce that $y$ is even. If $y=0$, we have the solution $(3,0)$. Otherwise, $y \geqslant 1$ and looking modulo 3 we have: $(-1)^{x} \equiv 1(\bmod 3)$ and thus $x$ is even.

Thus, we can rewrite the initial equation as $\left(2^{a}\right)^{2}-\left(3^{b}\right)^{2}=\left(2^{a}-3^{b}\right)\left(2^{a}+3^{b}\right)=7$, with $x=2 a, y=2 b$. In particular $2^{a}+3^{b}>2^{a}-3^{b}$ and $2^{a}+3^{b}>0$ and so $2^{a}+3^{b}=7$ and $2^{a}-3^{b}=1$. Then $2^{a}=\frac{7+1}{2}=4$ i.e. $a=2, b=1$.

We then find $(3,0)$ and $(4,2)$.

We verify reciprocally that these two pairs work because $2^{3}-3^{0}=8-1=7$ and $2^{4}-3^{2}=16-9=7$