Example 7 (Question from the 12th "Hope Cup" Invitational Competition) The solution set of the equation $\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$ is
Solve for \{2\}. Reason: From the relationship between exponents and logarithms and the given equation $y=\log _{5}\left(3^{x}+4^{x}\right)=\log _{4}\left(5^{x}-3^{x}\right)$, we have $3^{x}+4^{x}=5^{y}$,
$$
5^{x}-3^{x}=4^{y} .
$$

By adding (1) and (2), we get $5^{y}+4^{y}=5^{x}+4^{x}$.
Since $f(t)=5^{t}+4^{t}$ is a monotonically increasing function on $\mathbf{R}$, from (3), we have $f(x)=f(y)$, thus $x=y$. Therefore, (1) and (2) both reduce to $3^{x}+4^{x}=5^{x}$, which is $\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}=1$.

Now consider the function $g(x)=\left(\frac{3}{5}\right)^{x}+\left(\frac{4}{5}\right)^{x}$, which is monotonically increasing on $\mathbf{R}$. Noting that $g(2)=1$, we conclude $x=2$.