Example 4 The function $f^{-1}(x)$ is the inverse function of $f(x)=\log _{2}\left(2^{x}+1\right)$.
(1) If the equation $f^{-1}(x)=m+f(x)$ has a solution for $x$ in $[1,2]$, find the range of real number $m$;
(2) If the inequality $f^{-1}(x)>m+f(x)$ has a solution for $x$ in $[1,2]$, find the range of real number $m$;
(3) If the inequality $f^{-1}(x)>m+f(x)$ holds for any $x \in[1,2]$, find the range of real number $m$.
Solve $f^{-1}(x)=\log _{2}\left(2^{x}-1\right)$.
(1) $m=\log _{2}\left(2^{x}-1\right)-\log _{2}\left(2^{x}+1\right), m=\log _{2}\left(1-\frac{2}{2^{x}+1}\right)=g(x), x \in[1, 2]$ when, $1-\frac{2}{2^{x}+1} \in\left[\frac{1}{3}, \frac{3}{5}\right]$, which means the range of real number $m$ is $\left[\log _{2} \frac{1}{3}, \log _{2} \frac{3}{5}\right]$.
(2) $m<\log _{2}\left(1-\frac{2}{2^{x}+1}\right)$ has a solution on $[1,2]$, then $m<g_{\text {max }}$, which means the range of real number $m$ is $\left(-\infty, \log _{2} \frac{3}{5}\right)$.
(3) $m<\log _{2}\left(1-\frac{2}{2^{x}+1}\right)$ holds for any $x \in[1,2]$, then $m<g_{\text {min }}$, which means the range of real number $m$ is $\left(-\infty, \log _{2} \frac{1}{3}\right)$.