Example 3 Solve the system of equations
$$
\left\{\begin{array}{l}
\left(1+4^{2 x-y}\right) 5^{1-2 x+3}=1+2^{2 x-y+1}, \\
y^{3}+4 x+1+\ln \left(y^{2}+2 x\right)=0 .
\end{array}\right.
$$
(1999, Vietnam Mathematical Olympiad)
Solution: In (1), let $u=2 x-y$, then
$$
f(u)=5 \times\left(\frac{1}{5}\right)^{u}+5 \times\left(\frac{4}{5}\right)^{u}-2 \times 2^{u}-1=0 \text {. }
$$

Since $f(u)$ is monotonically decreasing on $\mathbf{R}$, and $f(1)=0$, thus, $u=1$, i.e., $2 x=y+1$.
Substituting the above equation into (2) yields
$$
\begin{array}{l}
g(y)=y^{3}+2 y+3+\ln \left(y^{2}+y+1\right)=0 . \\
g^{\prime}(y)=\frac{(2 y+1)^{2}+3 y^{2}\left(y+\frac{1}{2}\right)^{2}+\frac{1}{4} y^{2}+2}{\left(y+\frac{1}{2}\right)^{2}+\frac{3}{4}}>0 .
\end{array}
$$

Therefore, $g(y)$ is a monotonically increasing function on $\mathbf{R}$, and $g(-1)=0$. So, $y=-1$. Thus, $x=\frac{1}{2}(y+1)=0$.
Upon verification, $x=0, y=-1$ is the solution to the original system of equations. Note: The key to solving this problem is to construct the monotonically decreasing function $f(u)$ and the monotonically increasing function $g(y)$.