Finding all the numbers that fit $x! + y! = z!$

If $x, y \in \{0,1\}$, then we can always find a solution $z \in \{0, 1, 2\}$. The rest of the post will show that there are no other solutions.

Let us assume $y \geq x \geq 2$ without loss of generality.

Dividing both sides by $x!$ gives $$ 1 + y(y-1)\cdots(x+1) = z(z-1)\cdots(x+1). $$ If $y > x$, we see $x+1$ divides the right-hand side but not the left-hand side ($x+1$ divides one term in the sum but not the other), in which case there are no solutions.

If $y = x$, we may reduce the problem to that of solving $2y! = z!$. Since $y \geq 2$, the left-hand side always has more factors of $2$ than the right-hand side, in which case there are no solutions.

If $x>1$ and $ y \leq x$ then $$x!<x!+y! \leq 2x!<(x+1)!$$ therefore $$x!<z!<(x+1)!$$ so that the only solutions are $(x,y,z)=(0,0,2),(0,1,2),(1,0,2),(1,1,2)$.