Example 5 Find all $3<x<200(x \in Z)$, such that $x^{2}+(x+1)^{2}$ is a perfect square.
【Analysis】Inequality Estimation.
Let $x^{2}+(x+1)^{2}=y^{2}$.
By $(x, x+1)=1$, using the general solution formula of the Pythagorean equation, we get
$$
\left\{\begin{array}{l}
x=a^{2}-b^{2}, \\
x+1=2 a b, \text {, or }\left\{\begin{array}{l}
x+1=a^{2}-b^{2}, \\
y=a^{2}+b^{2}
\end{array} \quad\right. \text { ab, } \\
y=a^{2}+b^{2},
\end{array}\right.
$$

where, $a>b>0$, and $a, b$ are one odd and one even, $(a, b)=1$.
(1) From the first set of equations, we get
$$
b^{2}+2 a b-a^{2}=1 \Rightarrow(a-b)^{2}=2 b^{2}-1 \text {. }
$$

From $x+1=2 a b$, we get
$$
2 b^{2}<201 \Rightarrow b<10 \text {. }
$$

By verifying one by one, we know that $b=5, a=12, x=119$ satisfies the condition.
(2) From the second set of equations, we get
$$
a^{2}-2 a b-b^{2}=1 \Rightarrow(a-b)^{2}=2 b^{2}+1 \text {. }
$$

From $x=2 a b$, we get
$$
2 b^{2}<200 \Rightarrow b<10 \text {. }
$$

By verifying one by one, we know that $b=2, a=5, x=20$ satisfies the condition.
In summary, $x=119$ or 20.