First, we prove that $x, y, z$ given by equations (6) and (7) are certainly primitive solutions of (1) and $2 \mid y$. It is easy to verify that for any $r, s$ (not necessarily satisfying (7)), $x, y, z$ given by equation (6) are certainly solutions of (1) and $2 \mid y$. From $r>s>0$, we know that these are positive solutions. From equation (6), we have
$$(x, z) \mid 2 r^{2}, \quad (x, z) \mid 2 s^{2}$$

From this, using Theorem 2 and Theorem 3 of Chapter 1, §4, we derive
$$(x, z) \mid (2 r^{2}, 2 s^{2}) = 2 (r^{2}, s^{2})$$

By the condition $(s, r)=1$ and Theorem 5 of Chapter 1, §4, we have $(r^{2}, s^{2})=1$, thus
$$(x, z) \mid 2$$

By the condition $2 \nmid r+s$, we know $2 \nmid x$, so it must be that $(x, z)=1$. This proves the desired conclusion.