jmc

Setting $x=0$ in the second equation, we get $$f(-1)=f(0{)}^2-2.$$Setting $x=-1$ in the second equation, we get $$f(0)=(f(-1)+1{)}^2-1.$$Let $a=f(0)$ and $b=f(-1)$; then $b=a^2-2$ and $a=(b+1{)}^2-1.$ Substituting $b=a^2-2,$ we get $$a=(a^2-1{)}^2-1.$$This simplifies to $a^4-2a^2-a=0,$ which factors as $a(a+1)(a^2-a-1)=0.$ The quadratic $a^2-a-1=0$ has no integer solutions, so $a=0$ or $a=-1.$

Suppose $f(0)=a=0.$ Then $f(-1)=-2.$ Setting $x=-1$ in the first equation, we get $$f(3)-f(-1)=12,$$so $f(3)=f(-1)+12=10.$ But setting $x=2$ in the second equation, we get $$f(3)=(f(2)-2{)}^2+2,$$so $(f(2)-2{)}^2=8.$ No integer value for $f(2)$ satisfies this equation.

Therefore, $f(0)=a=-1.$ Setting $x=1$ in the second equation, we get $$f(0)=(f(1)-1{)}^2-1,$$so $(f(1)-1{)}^2=0,$ which forces $f(1)=1.$

Hence, $(f(0),f(1))=\boxed{(-1,1)}.$ Note that the function $f(n)=n^2+n-1$ satisfies the given conditions.