(England 1996) Find the positive integers $x, y, z$ such that $2^{x}+3^{y}=$ $z^{2}$.
If $x=0$, we verify that necessarily $y=1, z=2$ and that if $y=0$, necessarily $x=3$ and $z=3$. We assume therefore $x, y \geq 1$. The sequence is close to exercise 22. Modulo 3, we see that $x$ is odd. Let's write $x=2 w$, so that

$$
3^{y}=z^{2}-2^{2 w}=\left(z-2^{w}\right)\left(z+2^{w}\right)
$$

The GCD of $z-2^{w}$ and $z+2^{w}$ is equal to the GCD of $z$ and $2^{w}$, which is 1. Thus $z-2^{w}=1$ and $z+2^{w}=3^{y}$. By subtracting these two equations, we get

$$
2^{w+1}=3^{y}-1
$$

If $y \neq 2$, then $y \geq 3$ and Zsigmondy's theorem ensures the existence of a prime number $p$ dividing $3^{y}-1=2^{w+1}$ but not $3^{1}-1=2$, which is absurd. If $y=2$, we find the solution $(x, y, z)=(4,2,5)$.