Example 17 (2006 Iran National Team Selection Exam) Let $n$ be a fixed natural number.
(1) Find all solutions to the equation $\sum_{k=1}^{n}\left[\frac{x}{2^{k}}\right]=x-1$;
(2) When $m$ is a known natural number, find the number of all solutions to the equation $\sum_{k=1}^{n}\left[\frac{x}{2^{k}}\right]=x-m$.
Let \( x = \sum_{i=0}^{\infty} c_{i} 2^{i} \left( c_{i} \in \{0,1\} \right) \), then
\[
\begin{aligned}
\sum_{k=1}^{\infty} \left[ \frac{x}{2^{k}} \right] & = \sum_{k=1}^{\infty} \sum_{i=k}^{\infty} c_{i} 2^{i-k} = \sum_{i=1}^{\infty} \sum_{k=1}^{i} c_{i} 2^{i-k} = \sum_{i=1}^{\infty} c_{i} \sum_{k=1}^{i} 2^{i-k} \\
& = \sum_{i=1}^{\infty} c_{i} \left( 2^{i} - 1 \right) = \sum_{i=0}^{\infty} c_{i} 2^{i} - \sum_{i=0}^{\infty} c_{i} = x - f(x),
\end{aligned}
\]
where \( f(x) \) represents the number of "1"s in the binary representation of \( x \).
Since \( \sum_{k=1}^{\infty} \left[ \frac{x}{2^{k}} \right] \geqslant \sum_{k=1}^{n} \left[ \frac{x}{2^{k}} \right] \), thus, \( f(x) = 1 \).
Let \( x = 2^{k} \), then \( \left[ \frac{2^{k}}{2^{n+1}} \right] = 0 \).
Thus, \( k \leqslant n \).
Therefore, \( x = 2^{k} (0 \leqslant k \leqslant n) \).
(2) Let \( x = 2^{n} y + \sum_{i=1}^{l} 2^{r_{i}} \), where \( 0 \leqslant r_{1} < r_{2} < \cdots < r_{l} \leqslant n-1 \), then
\[
x - \sum_{k=1}^{n} \left[ \frac{x}{2^{k}} \right] = x - \left( 2^{n} y - y + \sum_{i=1}^{l} 2^{r_{i}} - l \right) = y + l = m.
\]
For a specific \( l \), \( y = m - l \), where \( 0 \leqslant l \leqslant m \).
For a specific \( l \), the number of ways to choose \( r_{1}, r_{2}, \cdots, r_{l} \) is \( C_{n}^{l} \).
0
Therefore, the total number of solutions to the equation is \( \sum_{l=0}^{m} C_{n}^{l} \).