Ukrainian National Mathematical Olympiad

Let's consider two cases for $x$.

For odd $x$, we have
$$z^2 \equiv 2^{2n+1} + 21^y \equiv 2 + 0 \equiv 2 \pmod{3},$$
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. $z - 2^n = 1$, $z + 2^n = 21^y \Rightarrow 2^{n+1} = 21^y - 1 \Rightarrow 2^{n+1} \equiv 6 \pmod 7$, which contradicts the fact that $2^{n+1} \equiv 1, 2, 4 \pmod 7$.

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 $7^y - 3^y \equiv 0 \pmod 8$ 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, $7^m + 3^m \equiv 2 \pmod 4$ and it is greater than $4$, so it cannot be a power of $2$.