Can we even call this Diophantine?

$\begin{aligned} x+y&=xy\\ xy-x-y&=0\\ xy-x-y+1&=1\\ (x-1)(y-1)&=1\\ x-1=1&, y-1=1\\ (x,y)&= (2,2) \end{aligned}$

The other solution is $(0,0)$ which doesn't satisfy the "natural numbers" requirement.

By A.M. G.M. Inequality,

$xy\leq\frac{(x+y)^2}{4}=x+y$

Case 1: $\frac{(x+y)^2}{4}<x+y$

By simplifying, we get $x+y<4$ , since we can cancel $(x+y)$ from both sides, as it is a positive natural number.

Only combinations are $(1,1),(2,1),(1,2)$ , none of which satisfy the equation.

Case 2: $\frac{(x+y)^2}{4}=x+y$

Equality only holds when $x=y$ , in the A.M. G.M. Inequality

Thus, the only combination is $(2,2)$ , which satisfies the equation.

Thus, there is only $\boxed{1}$ solution.

It is given that: $x+y=xy$ taking $(modx)$ , we have $x|y$ . Taking $(mody)$ , we have $y|x$ . So, we have $x=y$ . Substituting in our original eqn, we have $x=y=0$ and $x=y=2$ . Since $x,y$ are natural, $x=y=2$ is the only solution. So, the number of solutions is $\boxed{1}$