Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation
$$
  20^m - 10m^2 + 1 = 19^n.
$$
1. We start with the given equation:
   \[
   20^m - 10m^2 + 1 = 19^n \quad \text{(1)}
   \]

2. Consider the equation modulo 3:
   \[
   20 \equiv 2 \pmod{3} \quad \text{and} \quad 19 \equiv 1 \pmod{3}
   \]
   Thus,
   \[
   20^m \equiv 2^m \pmod{3}
   \]
   and
   \[
   19^n \equiv 1 \pmod{3}
   \]
   Therefore, equation (1) becomes:
   \[
   2^m - 10m^2 + 1 \equiv 1 \pmod{3}
   \]
   Simplifying further:
   \[
   2^m - 10m^2 \equiv 0 \pmod{3}
   \]
   Since \(10 \equiv 1 \pmod{3}\), we have:
   \[
   2^m - m^2 \equiv 0 \pmod{3}
   \]
   This implies:
   \[
   2^m \equiv m^2 \pmod{3}
   \]

3. We analyze the possible values of \(m\) modulo 3:
   - If \(m \equiv 0 \pmod{3}\):
     \[
     2^0 \equiv 1 \quad \text{and} \quad 0^2 \equiv 0 \pmod{3}
     \]
     This does not satisfy \(2^m \equiv m^2 \pmod{3}\).

   - If \(m \equiv 1 \pmod{3}\):
     \[
     2^1 \equiv 2 \quad \text{and} \quad 1^2 \equiv 1 \pmod{3}
     \]
     This does not satisfy \(2^m \equiv m^2 \pmod{3}\).

   - If \(m \equiv 2 \pmod{3}\):
     \[
     2^2 \equiv 1 \quad \text{and} \quad 2^2 \equiv 1 \pmod{3}
     \]
     This satisfies \(2^m \equiv m^2 \pmod{3}\).

4. Therefore, \(m\) must be even. Let \(m = 2k\) for some integer \(k\).

5. Substitute \(m = 2k\) into equation (1):
   \[
   20^{2k} - 10(2k)^2 + 1 = 19^n
   \]
   Simplifying:
   \[
   20^{2k} - 40k^2 + 1 = 19^n \quad \text{(2)}
   \]

6. Consider the equation modulo 5:
   \[
   20 \equiv 0 \pmod{5} \quad \text{and} \quad 19 \equiv -1 \pmod{5}
   \]
   Thus,
   \[
   20^{2k} \equiv 0 \pmod{5}
   \]
   and
   \[
   19^n \equiv (-1)^n \pmod{5}
   \]
   Therefore, equation (2) becomes:
   \[
   0 - 40k^2 + 1 \equiv (-1)^n \pmod{5}
   \]
   Simplifying further:
   \[
   1 \equiv (-1)^n \pmod{5}
   \]
   This implies \(n\) must be even. Let \(n = 2l\) for some integer \(l\).

7. Substitute \(n = 2l\) into equation (2):
   \[
   20^{2k} - 40k^2 + 1 = 19^{2l}
   \]

8. We now test small values of \(k\) and \(l\):
   - For \(k = 0\):
     \[
     20^0 - 40(0)^2 + 1 = 19^{2l}
     \]
     \[
     1 = 19^{2l}
     \]
     This implies \(2l = 0\), so \(l = 0\). Thus, \((m, n) = (0, 0)\) is a solution.

   - For \(k = 1\):
     \[
     20^2 - 40(1)^2 + 1 = 19^{2l}
     \]
     \[
     400 - 40 + 1 = 19^{2l}
     \]
     \[
     361 = 19^{2l}
     \]
     \[
     19^2 = 19^{2l}
     \]
     This implies \(2l = 2\), so \(l = 1\). Thus, \((m, n) = (2, 2)\) is a solution.

9. We check if there are any other solutions by considering larger values of \(k\):
   - For \(k \geq 2\), \(20^{2k}\) grows much faster than \(40k^2\), making it unlikely to balance the equation \(20^{2k} - 40k^2 + 1 = 19^{2l}\).

10. Therefore, the only solutions are \((m, n) = (0, 0)\) and \((m, n) = (2, 2)\).

The final answer is \( \boxed{ (0, 0) } \) and \((2, 2)\).