Example 15 (1999 Vietnam Mathematical Olympiad) Solve the system of equations
$$
\left\{\begin{array}{l}
\left(1+4^{2 x-y}\right) \cdot 5^{1-2 x+y}=1+2^{2 x-y+1}, \\
y^{3}+4 x+1+\ln \left(y^{2}+2 x\right)=0 .
\end{array}\right.
$$
Let $u=2 x-y$ in (1), then
$$
f(u)=5 \cdot\left(\frac{1}{5}\right)^{u}+5 \cdot\left(\frac{4}{5}\right)^{u}-2 \cdot 2^{u}-1=0 .
$$

Since $f(u)$ is monotonically decreasing on $\mathbf{R}$ and $f(1)=0$, we have $u=1$, i.e., $2 x=y+1$.
Substituting the above equation into (2) yields $g(y)=y^{3}+2 y+3+\ln \left(y^{2}+y+1\right)=0$.
And $g^{\prime}(y)=\frac{(2 y+1)^{2}+3 y^{2}\left(y+\frac{1}{2}\right)^{2}+\frac{1}{4} y^{2}+2}{\left(y+\frac{1}{2}\right)^{2}+\frac{3}{4}}>0$,
thus $g(y)$ is a monotonically increasing function on $\mathbf{R}$, and $g(-1)=0$.
Therefore, $y=-1$. Hence, $x=\frac{1}{2}(y+1)=0$.
Upon verification, $x=0, y=-1$ is a solution to the original system of equations.