Find all possible non-negative integer solution \((x,y)\) of the following equation: $$ x! + 2^y = (x+1)! $$

We know, \begin{align*} n! &= n \times (n-1) \times \dots \times 3 \times 2 \times 1\\ &= n \times (n-1)! \end{align*} Given that, \begin{align*} x! + 2^y &= (x+1)! \\ ⇒ 2^y &= (x+1)! - x! \\ ⇒ 2^y &= (x+1) \times x! - x! \\ ⇒ 2^y &= x! \times \{ (x+1) - 1 \} \\ ⇒ 2^y &= x! \times x \tag{1} \\ \end{align*}

From equation \(1\), it is clear that \(x \ngtr 2 \). When \(x\) is greater than \(2\), the prime factorization of \(x\) includes at least one prime number that cannot be expressed as a power of \(2\).

Therefore, \((x, y) = \{(1, 0), (2, 2)\} \)