Theorem 1 is proved by contradiction. Suppose there is a solution $x y z \neq 0$. Let $x_{0}, y_{0}, z_{0}$ be such a solution, making $\left|x_{0} y_{0} z_{0}\right|$ the smallest. We will prove that in this case, there must be another solution $x_{1}, y_{1}, z_{1}$, such that $01$ (it has been pointed out that $\left|x_{0} y_{0}\right|=1$ must have $z_{0}=0$, so it is impossible). Therefore,
$$0 \neq\left|x_{1} y_{1} z_{1}\right|<\left|x_{0} y_{0} z_{0}\right|$$

This contradicts the minimality of $\left|x_{0} y_{0} z_{0}\right|$. Therefore, it is impossible.
(ii) The case where $3 \mid u_{0}$. Let $u_{0}=3 v_{0}$, equation (23) becomes
$$z_{0}^{3}=18 v_{0}\left(3 v_{0}^{2}+w_{0}^{2}\right)$$

From equation (22), we know $\left(3 v_{0}, w_{0}\right)=1, 2 \nmid 3 v_{0}+w_{0}$, so
$$\left(18 v_{0}, 3 v_{0}^{2}+w_{0}^{2}\right)=1$$

By Corollary 5 of Chapter 1, §5, we have
$$0 \neq 18 v_{0}=\tilde{t}^{3}, \quad 3 v_{0}^{2}+w_{0}^{2}=\tilde{s}^{3} .$$

Here, $\tilde{s}$ is in the form of Lemma 2. By Lemma 2, there must be $\tilde{\alpha}, \widetilde{\beta}$ such that
$$\tilde{s}=\widetilde{\alpha}^{2}+3 \widetilde{\beta}^{2}, \quad(\widetilde{\alpha}, 3 \widetilde{\beta})=1$$

and $\square$
$$w_{0}=\widetilde{\alpha}^{3}-9 \widetilde{\alpha}^{2}, \quad v_{0}=3 \widetilde{\alpha}^{2} \widetilde{\beta}-3 \widetilde{\beta}^{3}$$

So $\square$
$$0 \neq(\tilde{t} / 3)^{3}=2 \widetilde{\beta}(\widetilde{\alpha}+\widetilde{\beta})(\widetilde{\alpha}-\widetilde{\beta}) .$$

It is easy to prove that $2 \beta, \tilde{\alpha}+\widetilde{\beta}, \tilde{\alpha}-\widetilde{\beta}$ are pairwise coprime. Thus, by Corollary 5 of Chapter 1, §5, we have
$$2 \widetilde{\beta}=z_{2}^{3}, \quad \widetilde{\alpha}+\widetilde{\beta}=x_{2}^{3}, \quad \tilde{\alpha}-\widetilde{\beta}=y_{2}^{3} .$$

Clearly, $x_{2}, y_{2}, z_{2}$ are solutions to the indeterminate equation (1), i.e.,
$$x_{2}^{3}+y_{2}^{3}=z_{2}^{3}$$

But in this case, we have
$$\begin{aligned} 0 \neq\left|x_{2} y_{2} z_{2}\right|^{3} & =|\tilde{t} / 3|^{3}=\left|2 v_{0} / 3\right|=\left|2 u_{0} / 9\right| \\ & =\left|x_{0}+y_{0}\right| / 9 \leqslant\left|x_{0}^{3}+y_{0}^{3}\right| / 9 \\ & =\left|z_{0}\right|^{3} / 9<\left|x_{0} y_{0} z_{0}\right|^{3}, \end{aligned}$$

This contradicts the minimality of $\left|x_{0} y_{0} z_{0}\right|$. Therefore, it is also impossible. The theorem is completely proved.