Rewrite the equation as $(z - 2^y)(z + 2^y) = 21^x$. Denote by $d$ the largest common divisor of $z - 2^y$ and $z + 2^y$; then $d$ divides $(z + 2^y) - (z - 2^y)$, so $d \mid 2^{y+1}$. Since $d$ is a divisor of $z + 2^y$ and $z + 2^y$ divides $21^x$, it follows that $d \mid (2^{y+1}, 21^x)$, so $d = 1$.
Consequently, either $z - 2^y = 1$ and $z + 2^y = 21^x$, or $z - 2^y = 3^x$ and $z + 2^y = 7^x$.
The first case leads to $21^x - 1 = 2^{y+1}$, equality which cannot hold, because $5 \mid 21^x - 1$ and $5 \nmid 2^{y+1}$.
The second case implies $7^x - 3^x = 2^{y+1}$. We easily see that $x \neq 0$, and for $x = 1$ we get the solution $(1, 1, 5)$. Assume that $x \ge 2$.
If $x$ is odd, then
$$2^{y-1} = 7^{x-1} + 7^{x-2} \cdot 3 + 7^{x-3} \cdot 3^2 + \dots + 7 \cdot 3^{x-2} + 3^{x-1},$$
a contradiction, because the left-hand side is even and right-hand side is odd.
If $x$ is an even number, $x = 2s$, $s \ge 1$, then $49^s - 9^s = 2^{y+1}$, which is impossible modulo 5, since $5 \mid 49^s - 9^s$.