Find all possible non-negative integer solution $(x,y)$ of the following equation- $$x! + 2^y =(x+1)!$$
Note: $x!=x \cdot (x-1)!$ and $0!=1$. For example, $5! = 5\times 4\times 3\times 2\times 1 = 120$.
\((2, 1)\)

   - For \(x = 3\):
     \[
     2^y = 3 \cdot 2! = 3 \cdot 2 = 6
     \]
     \(6\) is not a power of 2. No solution.

   - For \(x \geq 3\):
     \[
     x! \text{ is not a power of 2 for } x \geq 3
     \]
     No solutions for \(x \geq 3\).

7. Conclude that the only non-negative integer solutions are:
   \[
   (1, 0) \text{ and } (2, 1)
   \]

The final answer is \(\boxed{(1,0), (2,1)}\).