jmc

First, we expand the left side of our equation to get $$x^2+3x+2=x+3.$$Then we subtract $x+3$ from both sides to get a quadratic equation in standard form: $$x^2+2x-1=0.$$This doesn't factor in an obvious way, so we apply the quadratic formula, which gives solutions of $$x=\frac{-(2)\pm \sqrt{(2{)}^2-4(1)(-1)}}{2(1)}=\frac{-2\pm \sqrt{8}}{2}.$$We can simplify this, dividing $2$ out of the numerator and denominator to obtain $$x=-1\pm \sqrt{2}.$$Thus, the integers $m$ and $n$ referred to in the problem are $m=-1$, $n=2$, and their sum is $-1+2=\boxed{1}$.