Let's find all the triples of natural numbers $(x, y, z)$ for which $3^{x}+4^{y}=5^{z}$.
If $x$ is a positive integer, then $3^{x}$ is divisible by 3, and $4^{y}$, when divided by 3, always leaves a remainder of 1. Thus, $5^{z}$, when divided by 3, must also leave a remainder of 1, which means that $z$ must be even, i.e., $z=2t$, where $t$ is a natural number.

The equation $3^{x}+4^{y}=5^{z}=5^{2t}$ can be rearranged as

$$
3^{x}=5^{2t}-2^{2y}=\left(5^{t}-2^{y}\right)\left(5^{t}+2^{y}\right)
$$

Both factors on the right side are integers, so each must be a non-negative integer power of 3. If $y=0$, then $5^{t}-2^{y}=5^{t}-1$ would be divisible by 4, and thus could not be a divisor of $3^{x}$. We will show that if $y \geq 1$, then $5^{t}-2^{y}$ and $5^{t}+2^{y}$ are relatively prime. Both numbers are odd, so their greatest common divisor, $d$, is also odd, but $d$ divides their difference, $\left(5^{t}+2^{y}\right)-\left(5^{t}-2^{y}\right)=2^{y+1}$, which is only possible if $d=1$.

The product of two relatively prime numbers can only be $3^{x}$ if the smaller one is exactly 1, i.e., $5^{t}-2^{y}=1$, and $5^{t}+2^{y}=3^{x}$.

From the form $5^{t}-1=2^{y}$, we can see that $t$ cannot be even, because in that case the left side would be divisible by $5^{2}-1=24$, and thus by 3. If, however, $t$ is odd, then $5^{t}-5=5\left(5^{t-1}-1\right)$ is divisible by 24, so

$$
5^{t}-1=\left(5^{t}-5\right)+4
$$

leaves a remainder of 4 when divided by 8. Therefore, only $5^{t}-1=4$, or $t=1, z=2t=2, y=2$, and from $5^{t}+2^{y}=3^{x}$, we get $x=2$ is possible.

In the missing case $x=0$, our given equation becomes: $5^{z}=4^{y}+1=2^{2y}+1$. We previously saw that the only solution in non-negative integers is $y=z=1$.

Thus, two triples of non-negative integers satisfy the conditions of the problem: $(2,2,2)$ and $(0,1,1)$.