17. (10 points) Given the function
$$
f(x)=\log _{4}\left(4^{x}+1\right)+k x(k \in \mathbf{R})
$$

is an even function.
(1) Find the value of $k$;
(2) If the equation $f(x)=\log _{4}\left(\frac{m}{2^{x}}-1\right)$ has a solution, find the range of real numbers $m$.
17. (1) From the fact that $f(x)$ is an even function, we know
$$
\begin{array}{l}
f(-x)=f(x) \\
\Rightarrow \log _{4}\left(4^{-x}+1\right)+k(-x) \\
\quad=\log _{4}\left(4^{x}+1\right)+k x \\
\Rightarrow 2 k x=\log _{4}\left(4^{-x}+1\right)-\log _{4}\left(4^{x}+1\right) \\
\quad=\log _{4}\left(\frac{1+4^{x}}{4^{x}} \times \frac{1}{4^{x}+1}\right)=\log _{4} 4^{-x}=-x .
\end{array}
$$

By the arbitrariness of $x$, we know $k=-\frac{1}{2}$.
(2) From (1), we get
$$
\begin{array}{l}
\log _{4}\left(4^{x}+1\right)-\frac{1}{2} x=\log _{4}\left(\frac{m}{2^{x}}-1\right) \\
\Rightarrow \log _{4} \frac{4^{x}+1}{2^{x}}=\log _{4}\left(\frac{m}{2^{x}}-1\right) \\
\Rightarrow m=4^{x}+2^{x}+1 . \\
\text { Let } 2^{x}=t(t>0) .
\end{array}
$$

Thus, $m=t^{2}+t+1=\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}>1$.