Six. (25 points) Given
$$
f(x)=\lg (x+1)-\frac{1}{2} \log _{3} x .
$$
(1) Solve the equation: $f(x)=0$;
(2) Find the number of subsets of the set
$$
M=\left\{n \mid f\left(n^{2}-214 n-1998\right) \geqslant 0, n \in \mathbf{Z}\right\}
$$

(Li Tiehan, problem contributor)
(1) For any $0 < x_1 < x_2$, we have $\frac{x_1 + 1}{x_2 + 1} > \frac{x_1}{x_2}$, thus $\lg \frac{x_1 + 1}{x_2 + 1} > \lg \frac{x_1}{x_2}$. Therefore,
$$
\begin{array}{l}
f(x_1) - f(x_2) > \lg \frac{x_1}{x_2} - \log \frac{x_1}{x_2} \\
= \lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9}.
\end{array}
$$

Since $0 < \frac{x_1}{x_2} < 1$, we have $\lg \frac{x_1}{x_2} < 0$. Therefore,
$$
\lg \frac{x_1}{x_2} - \frac{\lg \frac{x_1}{x_2}}{\lg 9} > \lg \frac{x_1}{x_2} - \lg \frac{x_1}{x_2} = 0.
$$

Hence, $f(x)$ is a decreasing function on $(0, +\infty)$.
Noting that $f(9) = 0$, we have:
When $x > 9$, $f(x) < f(9) = 0$.
When $x < 9$, $f(x) > f(9) = 0$.
Therefore, $f(x) = 0$ has and only has one root $x = 9$.
$$
\begin{array}{l}
\text{(2) From } f(n^2 - 214n - 1998) \geq 0 \\
\Rightarrow f(n^2 - 214n - 1998) \geq f(9). \\
\text{Then }\left\{\begin{array}{l}
n^2 - 214n - 1998 \leq 9, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
n^2 - 214n - 2007 \leq 0, \\
n^2 - 214n - 1998 > 0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
(n - 223)(n + 9) \leq 0, \\
(n - 107)^2 > 1998 + 107^2 = 13447 > 115^2
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n > 222 \text{ or } n < -8
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
-9 \leq n \leq 223, \\
n \geq 223 \text{ or } n \leq -9.
\end{array}\right. \\
\end{array}
$$

Thus, $n = 223$ or $n = -9$.
Hence, $M = \{-9, 223\}$.
Therefore, the number of subsets of $M$ is 4.