Example 4 Find all pairs of coprime positive integers $\langle x, y\rangle$, such that $x\left|\left(y^{2}+210\right), y\right|\left(x^{2}+210\right)$.
$$
\begin{array}{l}
\text { For any pair of } x, y \text { that satisfy the conditions, let } \\
\left\{\begin{array}{l}
x^{2}+210=a y, \\
y^{2}+210=b x,
\end{array}\right. \\
\text { where } a, b \in \mathbf{Z}_{+} \text {. } \\
\text { Then } x^{4}+420 x^{2}+210^{2}=a^{2} y^{2}=a^{2}(b x-210) \\
\Rightarrow x\left(a^{2} b-x^{3}-420 x\right)=210\left(a^{2}+210\right) \text {. } \\
\text { Since } x, y \text { are coprime, from equations (1) and (2) we know } \\
(x, 210)=1,(y, 210)=1 \text {. } \\
\text { Therefore, from equation (3) we get } \\
x \mid\left(a^{2}+210\right) \text {. } \\
\text { Similarly, } y \mid\left(b^{2}+210\right) \text {. (4) } \\
\text { Also, } x \text { and } a \text { must be coprime, and } y \text { and } b \text { must be coprime. Thus, the pairs of coprime } \\
\text { positive integers }(x, a) \text { and }(y, b) \text { are also solutions to the problem. } \\
\text { Let } x=x_{0}, y=x_{1}, a=x_{-1}, b=x_{2} \\
\text { Then equations (1) and (2) can be written as } \\
\left\{\begin{array}{l}
x_{0}^{2}+210=x_{-1} x_{1}, \\
x_{1}^{2}+210=x_{0} x_{2} .
\end{array}\right. \\
\text { From equations (4) and (5), we know there exist positive integers } x_{-1} \text { and } x_{3} \text { such that } \\
\left\{\begin{array}{ll}
x_{2}^{2}-1+210=x_{-2} x_{0}, & \text { (4) } \\
x_{2}^{2}+210=x_{1} x_{3} . & \text { (5) }
\end{array}\right. \\
\text { Continuing this process, we know that from any pair of positive integers } \\
\text { } x, y \text { (i.e., } x_{0}, x_{1} \text {) that satisfy the conditions, we can obtain an infinite sequence of positive integers: } \\
\cdots, x_{-3}, x_{-2}, x_{-1}, x_{0}, x_{1}, x_{2}, x_{3}, \cdots \\
\text { where any three consecutive terms satisfy } \\
\frac{x_{n}^{2}+210}{x_{n+1}}=x_{n-1}, \frac{x_{n+1}^{2}+210}{x_{n}}=x_{n+2} \text {. } \\
\text { Furthermore, for any three consecutive terms } x_{n-1}, x_{n}, x_{n+1} \text { in the sequence, } \\
\text { the sum of the first and last terms divided by the middle term is a constant. } \\
\text { In fact, } \\
\frac{x_{n-1}+x_{n+1}}{x_{n}}=\frac{\left(x_{n}^{2}+210\right)+x_{n+1}^{2}}{x_{n} x_{n+1}} \\
=\frac{x_{n}^{2}+\left(x_{n+1}^{2}+210\right)}{x_{n} x_{n+1}}=\frac{x_{n}+x_{n+2}}{x_{n+1}} \\
\Rightarrow \frac{x_{n-1}+x_{n+1}}{x_{n}}=\frac{x_{-1}+x_{1}}{x_{0}}=\frac{a+y}{x} \text { (constant). } \\
\text { From equations (1) and (2), we get } x^{2}+b x=y^{2}+a y \text {. } \\
\text { Then } x+b=\frac{y+a}{x} y \text {. } \\
\text { Since } (x, y)=1 \text {, we have } x \mid(a+y) \text {, i.e., } t=\frac{a+y}{x} \\
\end{array}
$$

is an integer.
$$
\begin{array}{l}
\text { Also, } a+y \geqslant 2 \sqrt{a y}=2 \sqrt{x^{2}+210}>2 x, \text { so } \\
t=\frac{a+y}{x}>2 .
\end{array}
$$

Thus, the integer $t \geqslant 3$.
Therefore, the sequence satisfies the recurrence relation
$$
x_{n+1}=t x_{n}-x_{n-1}
$$

To find all such sequences, we need to determine $t$ and a pair of consecutive terms. Using equation (7), we can extend the sequence to the left and find all terms of the sequence (6), and any pair of consecutive terms in (6) is a solution to the problem.
From equation (7), we have
$$
\begin{array}{l}
\left(x_{n+1}, x_{n}\right)=\left(x_{n}, x_{n-1}\right)=\cdots \\
=\left(x_{1}, x_{0}\right)=(y, x)=1,
\end{array}
$$

which means any pair of consecutive terms in the sequence (6) are coprime, and each such sequence has a smallest positive integer.
To find all such sequences, we first consider the properties of sequence (2).
If $x_{n+1}>x_{n}$, then
$$
\begin{array}{l}
x_{n+2}=\frac{x_{n+1}^{2}+210}{x_{n}}>\frac{x_{n+1}^{2}}{x_{n}} \geqslant x_{n+1} ; \\
\text { If } x_{n-1}>x_{n}, \text { then } \\
x_{n-2}=\frac{x_{n-1}^{2}+210}{x_{n}}>\frac{x_{n-1}^{2}}{x_{n}} \geqslant x_{n-1} .
\end{array}
$$

Thus, each sequence (6) that satisfies the conditions starts with the smallest positive integer and strictly increases in both directions; or it starts with two equal smallest positive integers and strictly increases in both directions (when this happens, the two equal terms must both be 1, and the sequence is symmetric on both ends).
Let $x_{0}$ be the smallest positive integer in the sequence (6). Then
$x_{1} \geqslant x_{0}, x_{-1} \geqslant x_{0}$
Finally, we prove: $x_{0}210+x_{0}\left(x_{1}-x_{0}\right)$
$\Rightarrow x_{0}<x_{1}-x_{0}<\sqrt{210}$
$\Rightarrow x_{0} \leqslant 14$.
Furthermore, since $\left(x_{0}, 210\right)=1$, we have $x_{0} \in\{1,11,13\}$.
Also, from $x_{0}^{2}+210=x_{1} x_{-1}$, assume $x_{1} \leqslant x_{-1}$
Then $x_{1} \leqslant \sqrt{x_{0}^{2}+210}$
If $x_{0}=1$, then
$t=\frac{x_{0}^{2}+x_{1}^{2}+210}{x_{0} x_{1}}=\frac{x_{1}^{2}+211}{x_{1}}$.
Since $x_{1} \mid 211$ and $x_{1} \leqslant \sqrt{1^{2}+210}$, we have $x_{1}=1$.
In this case, $t=212$.
If $x_{0}=11,13$, there is no $x_{1}$ such that $t$ is a positive integer.
Thus, only $x_{0}=x_{1}=1, t=212$.
By symmetry, we only need to consider extending to the right.
From $x_{n+1}=212 x_{n}-x_{n-1}, x_{0}=x_{1}=1$, we get that each pair of consecutive terms $\left(x_{n}, x_{n+1}\right)(n \in \mathbf{N})$ in the sequence is a solution to the problem, where
$$
\begin{aligned}
x_{n}= & \frac{1}{107+\sqrt{11235}}(106+\sqrt{11235})^{n}+ \\
& \frac{1}{107-\sqrt{11235}}(106-\sqrt{11235})^{n}(n \in \mathbf{N}) .
\end{aligned}
$$
$$