jmc

By AM-GM, $$\sqrt{(xy+z)(xz+y)}\le \frac{(xy+z)+(xz+y)}{2}=\frac{xy+z+xz+y}{2}=\frac{(x+1)(y+z)}{2}.$$Again by AM-GM, $$\sqrt{(x+1)(y+z)}\le \frac{(x+1)+(y+z)}{2}=2,$$so $(x+1)(y+z)\le 4,$ and $$(xy+z)(xz+y)\le 4.$$Equality occurs when $x=y=z=1,$ so the maximum value is $\boxed{4}.$