Example 2 Given the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=a_{2}=a_{3}=1, a_{n+3} a_{n}=a_{n+1} a_{n+2}+1 \text {. }
$$

Find $a_{2 n+3} a_{2 n-1}-a_{2 n+1}^{2}$.
Analysis: This recursive sequence is non-homogeneous and nonlinear. To convert it into a homogeneous linear recursive sequence, we can eliminate the constant by taking the difference after one step of recursion to make it homogeneous, and then use a substitution to make it linear.
Solution: \( a_{n+3} a_{n} = a_{n+1} a_{n+2} + 1 \),
$$
a_{n+4} a_{n+1} = a_{n+2} a_{n+3} + 1 \text{. }
$$
(2) - (1) gives
$$
a_{n+4} a_{n+1} - a_{n+3} a_{n} = a_{n+2} a_{n+3} - a_{n+1} a_{n+2},
$$

which simplifies to
$$
\begin{array}{l}
a_{n+4} a_{n+1} + a_{n+1} a_{n+2} = a_{n+2} a_{n+3} + a_{n+3} a_{n}. \\
\text{Thus, } a_{n+1}(a_{n+4} + a_{n+2}) = a_{n+3}(a_{n+2} + a_{n}).
\end{array}
$$

Therefore, \(\frac{a_{n+1}}{a_{n+2} + a_{n}} = \frac{a_{n+3}}{a_{n+4} + a_{n+2}}\).
$$
\begin{array}{l}
\text{Let } b_{n} = \frac{a_{n+1}}{a_{n+2} + a_{n}}, \text{ then } b_{n} = b_{n+2}. \\
b_{1} = \frac{a_{2}}{a_{3} + a_{1}} = \frac{1}{2}, a_{4} = 2, b_{2} = \frac{a_{3}}{a_{4} + a_{2}} \\
= \frac{1}{3}, \text{ so, } \\
b_{1} = b_{3} = b_{5} = \ldots = \frac{1}{2}, b_{2} = b_{4} = b_{6} = \ldots = \frac{1}{3}. \\
\text{Thus, when } n \text{ is odd, } b_{n} = \frac{a_{n+1}}{a_{n+2} + a_{n}} = \frac{1}{2}, \\
\text{then } a_{n+2} = 2 a_{n+1} - a_{n}; \\
\text{when } n \text{ is even, } b_{n} = \frac{a_{n+1}}{a_{n+2} + a_{n}} = \frac{1}{3}, \text{ then } \\
a_{n+2} = 3 a_{n+1} - a_{n} \\
\text{Therefore, } a_{2 n+3} a_{2 n-1} - a_{2 n+1}^{2} \\
= (2 a_{2 n+2} - a_{2 n+1}) a_{2 n-1} - a_{2 n+1} (2 a_{2 n} - a_{2 n-1}) \\
= 2 a_{2 n+2} a_{2 n-1} - 2 a_{2 n} a_{2 n+1} \\
= 2 (a_{2 n+2} a_{2 n-1} - a_{2 n} a_{2 n+1}) \\
= 2 \times 1 = 2 \\
\end{array}
$$