Let's determine those positive integer triples $(x, y, z)$ for which $4^{x}+4^{y}+4^{z}$ is a perfect square.
Suppose $x \geq y \geq z$. Divide the sum $4^{z}=\left(2^{z}\right)^{2}$. Thus, $4^{x}+4^{y}+4^{z}$ is a perfect square if and only if $4^{x-z}+4^{y-z}+1$ is a perfect square. Let $x-z=a, y-z=b$, where $a \geq b$. If $a=b=0$, then the sum is 3, which is not a solution, so we can assume that $a>0$. Examine when there exists an integer $k$ such that $4^{a}+4^{b}+1=k^{2}$, that is, $4^{b}\left(4^{a-b}+1\right)=(k+1)(k-1)$.

1. If $a-b=b-1$, that is, $2 b-1=a$, then $4^{a}+4^{b}+1=4^{2 b-1}+4^{b}+1=\left(2^{2 b-1}+1\right)^{2}$. Thus, the sum is a perfect square when $2 b-1=a$.
2. If $a-b>b-1$, then $4^{a}+4^{b}+1$ is not a perfect square because the difference between $2 \cdot 4^{b-1}$ and $2 \cdot\left(4^{a-b}+1\right)$ is more than 2. Therefore, the difference between the two numbers is greater than 2, which does not match.
3. Finally, if $a-b>b-1$, that is, $2 b-1<a$, then $4^{a}+4^{b}+1$ is not a perfect square because it falls between two consecutive perfect squares:

$$
(2 a)^{2}=4^{a}<4^{a}+4^{b}+1=4^{a}+2 \cdot 2^{2 b-1}+1<4^{a}+2 \cdot 2^{a}+1=\left(2^{a}+1\right)^{2}
$$

Thus, $4^{a}+4^{b}+1$ is a perfect square if and only if $2 b-1=a$, that is, $x=2 b-1+z, y=b+z$, where $b$ and $z$ are arbitrary positive integers.

Based on the work of Zsolt Katona (Fazekas M. Főv. Gyak. Gimn., III. o.t.)