jmc

The given equation rewrites as $n^2=(x+y+z+1{)}^2+(x+y+z+1)-8$. Writing $r=x+y+z+1$, we have $n^2=r^2+r-8$. Clearly, one possibility is $n=r=\boxed{8}$, which is realized by $x=y=1,z=6$. On the other hand, for $r>8$, we have $r^2<r^2+r-8<(r+1{)}^2.$