International Mathematical Arhimede Contest

It is obvious that $x \ge 1$, $y$ is odd and $21^x \equiv 0 \pmod 7$. Since $2^3 \equiv 1 \pmod 7$ then we have $2^{3n} \equiv 1 \pmod 7$, $2^{3n+1} \equiv 2 \pmod 7$, and $2^{3n+2} \equiv 4 \pmod 7$. On the other hand, $y^3 \equiv 0 \pmod 7$ or $y^3 \equiv \pm 1 \pmod 7$. So, on account of the preceding, we have $x = 3n$ for all positive integer $n \ge 1$. Putting $x = 3n$ the first equation becomes
$$2^{3n} + 21^{3n} = y^3 \Leftrightarrow (y - 21^n)(y^2 + y \cdot 21^n + 21^{2n}) = 2^{3n}$$
Form the preceding immediately follows that $(y^2 + y \cdot 21^n + 21^{2n}) \mid 2^{3n}$ and this it is not possible on account that $y^2 + y \cdot 21^n + 21^{2n} > 1$ is an odd positive integer. This completes the proof of the first part of the statement.
To find the solutions in positive integers of the second equation we distinguish the following cases:

1. When $x = 0$ we have $(z - 1)(z + 1) = 21^y$. Let $d = \gcd(z - 1, z + 1)$. Since $d|21^y$ and $d|(z + 1) - (z - 1) = 2$, then $d = 1$ and
$$\begin{cases} z - 1 = 1, \\ z + 1 = 21^y \end{cases} \Rightarrow 21^y = 3$$
which is impossible.

2. When $x = 1$ we have $2 + 21^y = z^2$ which it is not possible because $2 + 21^y \equiv 2 \pmod 3$ and $z^2 \equiv 0 \pmod 3$ or $z^2 \equiv 1 \pmod 3$.

3. When $x = 2$ we have $(z - 2)(z + 2) = 21^y$. Let $d = \gcd(z - 2, z + 2)$. Since $d|21^y$ and $d|((z + 2) - (z - 2) = 4)$, then $d = 1$ and we have
$$\begin{cases} z - 2 = 1, \\ z + 2 = 21^y \end{cases} \quad \text{or} \quad \begin{cases} z - 2 = 3^y, \\ z + 2 = 7^y \end{cases}$$
In the first case we obtain, after subtraction, $21^y = 5$ (impossible). In the second case, we get $7^y - 3^y = 4$. For $y \ge 2$ we have
$$7^y - 3^y = 3^y \left[ \left( \frac{7}{3} \right)^y - 1 \right] \ge 9 \left[ \left( \frac{7}{3} \right)^2 - 1 \right] > 4,$$
and $y = 1$ verifies the equation $2^2 + 21 = 5^2$.
When $y = 0$ we have the equation $(z - 1)(z + 1) = 2^x$. Let $d = (z - 1, z + 1)$. Since $d|2^x$ and $d|(z + 1) - (z - 1) = 2$ then $d = 2$ because $z$ is odd. So, we have
$$\begin{cases} z - 1 = 2, \\ z + 1 = 2^{x-1} \end{cases} \Rightarrow z = 3, x = 3$$
for which is $2^3 + 21^0 = 3^2$.

4. Finally, assume that $x \ge 3$ and $y \ge 1$ and observe that $z$ is odd. We prove that $y$ is even number. Indeed, if $y = 2p+1$, then on account that $21^{2p} \equiv 1 \pmod 8$ and $21 \equiv 5 \pmod 8$ we have $21^{2p+1} \equiv 5 \pmod 8$. So, $2^x + 21^y \equiv 5 \pmod 8$ and $z^2 \equiv 1 \pmod 8$ (impossible). Therefore, $y = 2p$ and then we have $2^x = z^2 - 21^{2p} = (z - 21^p)(z + 21^p)$. Let $d = (z - 21^p, z + 21^p)$. Since $d|2^x$, $z$ is odd and $d|(z + 21^p) - (z - 21^p) = 2 \cdot 21^p$, we have
$$\begin{cases} z - 21^p = 2, \\ z + 21^p = 2^{x-1} \end{cases}$$
from which follows $2^{x-1} - 2 = 2 \cdot 21^p$ or $2^{x-2} = 1 + 21^p$. When $x = 3$, we have $p = 0$. That is, $y = 0$ in contradiction with $y \ge 1$. For $x \ge 4$, we have $2^{x-2} \equiv 0 \pmod 4$ and $(1 + 21^p) \equiv 2 \pmod 4$.
In conclusion, the solutions are $x = 3, y = 0, z = 3$ and $x = 2, y = 1, z = 5$.