Find all values of $b$ such that the difference between the maximum and minimum values of $f(x) = x^2-2bx-1$ on the interval $[0, 1]$ is $1$.
1. Consider the function \( f(x) = x^2 - 2bx - 1 \) on the interval \([0, 1]\).
2. To find the maximum and minimum values of \( f(x) \) on \([0, 1]\), we first evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 1 \), and at any critical points within the interval.
3. The critical points are found by setting the derivative \( f'(x) \) to zero:
   \[
   f'(x) = 2x - 2b
   \]
   Setting \( f'(x) = 0 \):
   \[
   2x - 2b = 0 \implies x = b
   \]
4. The critical point \( x = b \) is within the interval \([0, 1]\) if \( 0 \leq b \leq 1 \).

5. Evaluate \( f(x) \) at the endpoints and the critical point:
   \[
   f(0) = 0^2 - 2b \cdot 0 - 1 = -1
   \]
   \[
   f(1) = 1^2 - 2b \cdot 1 - 1 = 1 - 2b - 1 = -2b
   \]
   \[
   f(b) = b^2 - 2b \cdot b - 1 = b^2 - 2b^2 - 1 = -b^2 - 1
   \]

6. We need the difference between the maximum and minimum values of \( f(x) \) on \([0, 1]\) to be 1. Consider two cases:

   - **Case 1: \( b \in [0, 1] \)**
     - Minimum value: \( f(b) = -b^2 - 1 \)
     - Maximum value: \( f(0) = -1 \) or \( f(1) = -2b \)
     - If the maximum value is \( f(0) = -1 \):
       \[
       -1 - (-b^2 - 1) = b^2 = 1 \implies b = \pm 1
       \]
       Since \( b \in [0, 1] \), \( b = 1 \).
     - If the maximum value is \( f(1) = -2b \):
       \[
       -2b - (-b^2 - 1) = b^2 - 2b + 1 = (b - 1)^2 = 1 \implies b - 1 = \pm 1
       \]
       Solving \( b - 1 = 1 \) gives \( b = 2 \) (not in \([0, 1]\)).
       Solving \( b - 1 = -1 \) gives \( b = 0 \).

   - **Case 2: \( b \notin [0, 1] \)**
     - Minimum and maximum values occur at the endpoints \( x = 0 \) and \( x = 1 \):
       \[
       f(0) = -1, \quad f(1) = -2b
       \]
       \[
       |-1 - (-2b)| = 1 \implies |-1 + 2b| = 1
       \]
       Solving \( -1 + 2b = 1 \) gives \( 2b = 2 \implies b = 1 \).
       Solving \( -1 + 2b = -1 \) gives \( 2b = 0 \implies b = 0 \).

7. Therefore, the values of \( b \) that satisfy the condition are \( b = 0 \) and \( b = 1 \).

The final answer is \( \boxed{ b = 0 } \) and \( b = 1 \).