jmc

Let $a=x-1$ and $b=y-1.$ Then $x=a+1$ and $y=b+1,$ so $$\begin{array}{rl}\frac{x^2}{y-1}+\frac{y^2}{x-1}& =\frac{(a+1{)}^2}{b}+\frac{(b+1{)}^2}{a}\\ & =\frac{a^2+2a+1}{b}+\frac{b^2+2b+1}{a}\\ & =2\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{a^2}{b}+\frac{1}{b}+\frac{b^2}{a}+\frac{1}{a}.\end{array}$$By AM-GM, $$\frac{a}{b}+\frac{b}{a}\ge 2\sqrt{\frac{a}{b}\cdot \frac{b}{a}}=2$$and $$\frac{a^2}{b}+\frac{1}{b}+\frac{b^2}{a}+\frac{1}{a}\ge 4\sqrt[4]{\frac{a^2}{b}\cdot \frac{1}{b}\cdot \frac{b^2}{a}\cdot \frac{1}{a}}=4,$$so $$2\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{a^2}{b}+\frac{1}{b}+\frac{b^2}{a}+\frac{1}{a}\ge 2\cdot 2+4=8.$$Equality occurs when $a=b=1,$ or $x=y=2,$ so the minimum value is $\boxed{8}.$