Among the three numbers $x y+1, x y-x-y+3$ and $-2 x y+x+y+2$, we denote $K$ as the smallest, $L$ as the second smallest, and $M$ as the largest. Therefore, $K \leqslant L \leqslant M$ and

$$3 M \geqslant K+L+M=(x y+1)+(x y-x-y+3)+(-2 x y+x+y+2)=6,$$

which means that $M \geqslant 2$.
Furthermore, if $M=2$, the inequalities $K \leqslant L \leqslant M$ are in fact equalities, which means that

$$x y+1=x y-x-y+3=-2 x y+x+y+2=2$$

Let $p$ be the product $x y$ and $s$ be the sum $x+y$. The equalities in equation (1) are satisfied if and only if $p=1$ and $s=2$.
The arithmetic-geometric mean inequality generally indicates that

$$\frac{s^{2}}{4}=\left(\frac{x+y}{2}\right)^{2} \geqslant x y=p$$

with equality if and only if $x=y$. This result can also be seen if we note that

$$\left(\frac{x+y}{2}\right)^{2}=\left(\frac{x-y}{2}\right)^{2}+x y$$

Here, since we want to have $s^{2} / 4=1=p$, we must also have $x=y$, so that $x=y=1$.
Finally, conversely, it is easily verified that if $x=y=1$,

$$x y+1=x y-x-y+3=-2 x y+x+y+2=2$$

so that $M=2$ as well.