Nice exercise. Check that $xy+4 - 2(x+y) = (x-2)(y-2)>0$ yields, together with $x>2$, that $y>2$. So, $xy>4$. Now assume $x+y\ge xy$. Then, $xy+4>2(x+y)\ge 2xy$, yielding $4>xy$, a contradiction.

Or, since, as proven above, both numbers are greater than 2, we have $$(x-1)(y-1) > (2-1)(2-1)=1 \Rightarrow xy > x+y$$