62nd Ukrainian National Mathematical Olympiad

Let's consider two cases for $x$.

For odd $x$, we have $$z^2\equiv 2^{2n+1}+21^y\equiv 2+0\equiv 2(\mathrm{mod}3),$$ which has no solutions.

Let $x=2n$. Then $$21^y=z^2-2^{2n}=(z-2^n)(z+2^n).$$ Suppose that one of the two prime numbers $3$ or $7$ divides each of the two factors on the right-hand side. Then this prime also divides the number $2z=(z+2^n)+(z-2^n)$ and hence $z$. However, this contradicts the equation $2^x=z^2-21^y$, in which the right-hand side is divisible by the corresponding prime number, while the left-hand side is not. Therefore, these factors are coprime. Hence, we have the following two cases.

Case 1. $z-2^n=1$, $z+2^n=21^y\Rightarrow 2^{n+1}=21^y-1\Rightarrow 2^{n+1}\equiv 6(\mathrm{mod}7)$, which contradicts the fact that $2^{n+1}\equiv 1,2,4(\mathrm{mod}7)$.

Case 2. $z-2^n=3^y$, $z+2^n=7^y\Rightarrow z=2^n+3^y$ and $2^{n+1}=7^y-3^y$. For $y=1$, we find that $x=2$ and $z=5$, which satisfies the equation. For $y\ge 2$, the expression $7^y-3^y\equiv 0(\mathrm{mod}8)$ is only possible for even $y$. If $y=2m$, then $2^{n+1}=(7^m-3^m)(7^m+3^m)$. But under these conditions, $7^m+3^m\equiv 2(\mathrm{mod}4)$ and it is greater than $4$, so it cannot be a power of $2$.