Problem 4. Solve in the set of natural numbers the equation $21^{x}+4^{y}=z^{2}$.
Solution. We have $\left(z-2^{y}\right)\left(z+2^{y}\right)=21^{x}$ and, denoting $d=\left(z-2^{y}, z+2^{y}\right)$, it follows that $d \mid\left(z+2^{y}\right)-\left(z-2^{y}\right)$, so $d \mid 2^{y+1}$. Since $d \mid z+2^{y}$ and $z+2^{y} \mid 21^{x}$, it follows that $d \mid\left(2^{y+1}, 21^{x}\right)$, so $d=1$.

As a result, we need to analyze two cases: $z-2^{y}=1$ and $z+2^{y}=21^{x}$, or $z-2^{y}=3^{x}$ and $z+2^{y}=7^{x}$.

In the first case, we get $21^{x}-1=2^{y+1}$, an equality that cannot hold, because $5 \mid 21^{x}-1$ and $5 \nmid 2^{y+1}$.

In the second case, we have $7^{x}-3^{x}=2^{y+1}$, for which we observe that $x=0$ does not work, and $x=1$ leads to the solution $(1,1,5)$. We continue to study the case $x \geq 2$.

If $x$ is odd, we can write:

$$
2^{y-1}=7^{x-1}+7^{x-2} \cdot 3+7^{x-3} \cdot 3^{2}+\ldots+7 \cdot 3^{x-2}+3^{x-1}
$$

a contradiction, because the right-hand side is a sum of an odd number of odd numbers.

For $x$ even, $x=2 s, s \geq 1$, we get $49^{s}-9^{s}=2^{y+1}$, an equality that cannot hold because $5 \mid 49^{s}-9^{s}$.

In conclusion, the only solution is $(1,1,5)$.