Ta có : $xy-x+2y=3$
$⇔x.(y-1)+2.(y-1)=1$
$⇔(y-1).(x+2)=1$
Do $x,y$ nguyên nên $y-1,x+2$ nguyên
Ta xét các Trường hợp :
TH1 : $\left\{\begin{array}{l}y-1=1\\x+2=1\end{array} \right.$
$⇔\left\{\begin{array}{l}y=2\\x=-1\end{array} \right.$
TH2 : $\left\{\begin{array}{l}y-1=-1\\x+2=-1\end{array} \right.$
$⇔\left\{\begin{array}{l}y=0\\x=-3\end{array} \right.$
Vậy : $(x,y) ∈ ${$(-1,2);(-3,0)$}