That's not an Identity!

Since our expression is symmetric and both x and y are positive integers, we can assume, that 1 ≤ y ≤ x .

We will show, that for x > 1, (x + y)! > x! + y!

As (x + y)! ≥ (x+1)! and x! + x! ≥ x! + y! , therefore it is sufficient to show, that:

(x + 1)! > x! + x!

(x + 1)x! > 2x!

x + 1 > 2

x > 1.

This means, that the equation never holds for any positive integer, which is greater than 1 (the smallest positive integer), which leaves us with the only possible (x, y) pair of (1, 1).

And x = y = 1 is a solution, since (1+1)! = 2! = 2 = 1 + 1 = 1! + 1! .

Hence, our answer should be:

$x + y = 1 + 1 = \boxed {2}$

The numbers that satisfy $(x+y)! = x!+y!$ is only giving us the solution that $x = 1 , y = 1$ . Because $(1+1)! = 1! + 1!$ is always true. So $x + y$ is $1 + 1 = \boxed{2}$ .