Solve in natural integers the equation:

$$
4^{x}+3^{y}=z^{2}
$$
$4^{x}=\left(2^{x}\right)^{2}$ so the equation can be rewritten as

$$
3^{y}=\left(z-2^{x}\right)\left(z+2^{x}\right)
$$

Let $d$ be the greatest common divisor of the two factors on the right. Then, by taking the difference of the two factors, we deduce that $d$ divides $2^{x+1}$. However, we know that each factor is a power of 3. Therefore, the smallest factor is 1. So $z-2^{x}=1$ which means $z=2^{x}+1$. Substituting $\mathbf{z}$ into the previous equation, we get $2^{x+1}+1=3^{y}$. $(x=0, y=1, z=2)$ is a solution. We now focus on solutions with $x \geq 1$. By taking the equation modulo 4, we notice that $y$ is even. So $y=2 y_{1}$.

The equation can be rewritten as:

$$
2^{x+1}=\left(3^{y_{1}}-1\right)\left(3^{y_{1}}+1\right)
$$

$\operatorname{GCD}\left(\left(3^{y_{1}}-1\right),\left(3^{y_{1}}+1\right)\right) \mid 2$. Since both factors are even, the GCD is 2. Therefore, $3^{y_{1}}-1=2$ and $y_{1}=1$, so $(y=2, x=2, z=5)$ is the second solution.