[Solution] Since $2^{24}+2^{24}=2^{25}$, we have
$$\left(2^{8}\right)^{3}+\left(2^{6}\right)^{4}=\left(2^{5}\right)^{5}$$

That is, $x=2^{8}=256 ; y=2^{6}=64, z=2^{5}=32$ is a set of solutions to the given equation.

On the other hand, if $\left(x_{0}, y_{0}, z_{0}\right)$ is a set of solutions to the given equation in the set of natural numbers, then $\left(k^{20} x_{0}, k^{15} y_{0}, k^{12} z_{0}\right)$ is also a solution, where $k$ is any natural number.

Therefore, the set of solutions to the given equation in the set of natural numbers is infinite.