Let $x, y, z$ be positive integer solutions of (16), satisfying $(x, y)=1$. Therefore, $x, y, z^{2}$ are primitive solutions of equation (1). By Lemma 1, $x, y$ are one odd and one even, without loss of generality, assume $2 \mid y$. By Theorem 2, we must have
$$x=r^{2}-s^{2}, \quad y=2 r s, \quad z^{2}=r^{2}+s^{2},$$

where $r, s$ satisfy equation (7). Thus, $r, s, z$ are also primitive solutions of equation (1). If $2 \mid s$, then by Theorem 2, we have
$$r=a^{2}-b^{2}, \quad s=2 a b, \quad z=a^{2}+b^{2},$$

where $a, b$ satisfy (note $r>s$)
$$a>b>0, \quad(a, b)=1, \quad 2 \nmid a+b, \quad a^{2}-b^{2}>2 a b .$$

From equations (20) and (21), we get
$$x=a^{4}+b^{4}-6 a^{2} b^{2}, \quad y=4 a b\left(a^{2}-b^{2}\right), \quad z=a^{2}+b^{2} .$$

From equation (22), we get
$$(\sqrt{2}-1) a>b>0, \quad(a, b)=1, \quad 2 \nmid a+b .$$

If $2 \mid r$, then by Theorem 2, we have
$$r=2 a b, \quad s=a^{2}-b^{2}, \quad z=a^{2}+b^{2},$$

where $a, b$ satisfy (note $r>s$)
$$a>b>0, \quad(a, b)=1, \quad 2 \nmid a+b, \quad 2 a b>a^{2}-b^{2} .$$

From equations (20) and (25), we get
$$x=6 a^{2} b^{2}-a^{4}-b^{4}, \quad y=4 a b\left(a^{2}-b^{2}\right), \quad z=a^{2}+b^{2} .$$

From equation (26), we get
$$a>b>(\sqrt{2}-1) a>0, \quad(a, b)=1, \quad 2 \nmid a+b .$$

From equations (23) and (27), and equations (24) and (28), we conclude that when $2 \mid y$, the solutions are given by equations (17) and (19). By symmetry, when $2 \mid x$, the solutions are given by equations (18) and (19). Additionally, it is easy to verify directly that the $x, y, z$ given by equations (17), (18), and (19) are indeed solutions of equation (16) satisfying $(x, y)=1$. The theorem is proved.