$F_n$ is the Fibonacci sequence $F_0 = F_1 = 1$, $F_{n+2} = F_{n+1} + F_n$. Find all pairs $m > k \geq 0$ such that the sequence $x_0, x_1, x_2, ...$ defined by $x_0 = \frac{F_k}{F_m}$ and $x_{n+1} = \frac{2x_n - 1}{1 - x_n}$ for $x_n \not = 1$, or $1$ if $x_n = 1$, contains the number $1$
1. **Define the Fibonacci sequence and the sequence \( x_n \):**
   - The Fibonacci sequence is defined as \( F_0 = F_1 = 1 \) and \( F_{n+2} = F_{n+1} + F_n \).
   - The sequence \( x_n \) is defined by \( x_0 = \frac{F_k}{F_m} \) and \( x_{n+1} = \frac{2x_n - 1}{1 - x_n} \) for \( x_n \neq 1 \), or \( x_{n+1} = 1 \) if \( x_n = 1 \).

2. **Assume \( x_n = 1 \) at some point:**
   - If \( x_n = 1 \), then \( x_{n+1} = 1 \) by the definition of the sequence.
   - We need to find pairs \( (k, m) \) such that \( x_n = 1 \) for some \( n \).

3. **Construct a new sequence \( y_i \):**
   - Define \( y_0 = 1 \) and \( y_{n+1} = \frac{y_n + 1}{y_n + 2} \).
   - If \( x_i = y_j \), then \( x_{i+1} = y_{j-1} \).

4. **Prove by induction that \( y_i = \frac{F_{2i}}{F_{2i+1}} \):**
   - Base case: For \( i = 0 \), \( y_0 = 1 = \frac{F_0}{F_1} = \frac{1}{1} \).
   - Inductive step: Assume \( y_i = \frac{F_{2i}}{F_{2i+1}} \). Then,
     \[
     y_{i+1} = \frac{y_i + 1}{y_i + 2} = \frac{\frac{F_{2i}}{F_{2i+1}} + 1}{\frac{F_{2i}}{F_{2i+1}} + 2} = \frac{\frac{F_{2i} + F_{2i+1}}{F_{2i+1}}}{\frac{F_{2i} + 2F_{2i+1}}{F_{2i+1}}} = \frac{F_{2i+2}}{F_{2i+3}}
     \]
     where \( F_{2i+2} = F_{2i} + F_{2i+1} \) and \( F_{2i+3} = F_{2i+1} + F_{2i+2} \).

5. **Identify pairs \( (k, m) \):**
   - We need \( x_0 = \frac{F_k}{F_m} \) to be one of the \( y_i \) values.
   - From the induction, \( y_i = \frac{F_{2i}}{F_{2i+1}} \). Therefore, \( x_0 = \frac{F_k}{F_m} = \frac{F_{2i}}{F_{2i+1}} \).
   - This implies \( k = 2i \) and \( m = 2i+1 \).

6. **Prove uniqueness of the pairs \( (k, m) \):**
   - Assume there exist \( k' \) and \( m' \) such that \( F_{k'} \cdot F_{2i+1} = F_{m'} \cdot F_{2i} \).
   - Using the properties of Fibonacci numbers, \( F_x = F_2 \cdot F_x < F_4 \cdot F_{x-2} < \ldots < F_1 \cdot F_{x+1} = F_{x+1} \).
   - This implies there are no other pairs \( (k, m) \) such that \( F_i \cdot F_j = F_k \cdot F_l \) for \( 2 \leq i, j, k, l \).

\(\blacksquare\)

The final answer is \( \boxed{ (k, m) = (2i, 2i+1) } \) for \( i \geq 0 \).