jmc

Squaring the equation $x+y=2,$ we get $x^2+2xy+y^2=4.$ Also, $4xy-4z^2=4,$ so $$x^2+2xy+y^2=4xy-4z^2.$$Then $x^2-2xy+y^2+4z^2=0,$ which we write as $$(x-y{)}^2+4z^2=0.$$For this equation to hold, we must have $x=y$ and $z=0,$ and if $x=y,$ then $x=y=1.$

Therefore, there is only $\boxed{1}$ solution, namely $(x,y,z)=(1,1,0).$