4. $\frac{8}{9}<m \leqslant 1$.

Obviously, $f(x)$ is monotonically increasing, and $f(x)=1$ if and only if $x=0$.
Let $t=\log _{2} x$. Then
$$
f\left(2 t^{2}-2 t+\frac{4}{m}-4\right)=1 \text {. }
$$

Therefore, $2 t^{2}-2 t+\frac{4}{m}-4=0$ has two real solutions in $t \in\left[0, \frac{3}{2}\right]$.
Let $g(t)=2 t^{2}-2 t+\frac{4}{m}-4$. Then
$$
\left\{\begin{array}{l}
g(0) \geqslant 0, \\
g\left(\frac{1}{2}\right)<0, \\
g\left(\frac{3}{2}\right) \geqslant 0
\end{array} \Rightarrow \frac{8}{9}<m \leqslant 1 .\right.
$$
4. Given the function
$$
f(x)=\log _{2}\left(4^{x}+1\right)+m x(m>0)
$$

The equation
$$
f\left(8\left(\log _{4} x\right)^{2}+2 \log _{2} \frac{1}{x}+\frac{4}{m}-4\right)=1
$$

has exactly two distinct real solutions in the interval $[1,2 \sqrt{2}]$. Then the set of all possible values of $m$ is