Four. (20 points) Given the function $f_{1}(x)=\sqrt{x^{2}+48}$, when $n \in \mathbf{Z}_{+}, n \geqslant 2$,
$$
f_{n}(x)=\sqrt{x^{2}+6 f_{n-1}(x)} \text {. }
$$

Find the real solutions to the equation $f_{n}(x)=2 x$.
Because $f_{n}(x)=\sqrt{x^{2}+6 f_{n-1}(x)}>0$, the real solution of the equation $f_{n}(x)=2 x$ must be positive.
First, prove that $x=4$ is a solution of the equation $f_{n}(x)=2 x$.
Use mathematical induction.
(1) When $n=1$, $f_{1}(x)=\sqrt{x^{2}+48}=2 x$. Squaring both sides gives $x^{2}=16$.
Since $x>0$, we have $x=4$.
Thus, when $n=1$, the solution of the equation $f_{n}(x)=2 x$ is $x=4$.
(2) Assume that when $n=k$, $x=4$ is a solution of the equation $f_{k}(x)=2 x$, i.e., $f_{k}(4)=8$.
Then when $n=k+1$,
$$
\begin{array}{l}
f_{k+1}(4)-2 \times 4=\sqrt{4^{2}+6 f_{k}(4)}-8 \\
=\sqrt{16+6 \times 8}-8=0 .
\end{array}
$$

Therefore, $x=4$ is also a solution of the equation $f_{k+1}(x)=2 x$.
In summary, $x=4$ is a solution of the equation $f_{n}(x)=2 x$ for $n \in \mathbf{Z}_{+}$.
Next, prove that the equation $f_{n}(x)=2 x$ has no other solutions.
(1) If $x>4$, then
$$
84$, 
$$
8<f_{n}(x)<2 x \text{. }
$$

Thus, the equation $f_{n}(x)=2 x$ has no real solutions greater than 4.
(2) When $0<x<4$, it can be proven that:
$$
2 x<f_{n}(x)<8 \text{. }
$$

Thus, the equation $f_{n}(x)=2 x$ has no solutions less than 4.
Therefore, the equation $f_{n}(x)=2 x$ has a unique solution $x=4$.