jmc

Setting $n=0,$ we get $$f(m)+f(-1)=f(m)f(0)+2.$$If $f(0)\ne 1,$ then $f(m)$ is equal to some constant, say $c.$ Then $$2c=c^2+2,$$which has no integer solutions. Therefore, $f(0)=1,$ and then $f(-1)=2.$

Setting $n=1,$ we get $$f(m+1)+f(m-1)=f(1)f(m)+2.$$Let $a=f(1)$; then $$f(m+1)=af(m)-f(m-1)+2.$$Since $f(0)=1$ and $f(1)=a,$ $$\begin{array}{rl}f(2)& =af(1)-f(0)+2=a^2+1,\\ f(3)& =af(2)-f(1)+2=a^3+2,\\ f(4)& =af(3)-f(2)+2=a^4-a^2+2a+1,\\ f(5)& =af(4)-f(3)+2=a^5-2a^3+2a^2+a.\end{array}$$Setting $m=n=2,$ we get $$f(4)+f(3)=f(2{)}^2+2.$$Then $(a^4-a^2+2a+1)+(a^3+2)=(a^2+1{)}^2+2,$ which simplifies to $$a^3-3a^2+2a=0.$$This factors as $a(a-1)(a-2)=0.$ Hence, $a\in \{0,1,2\}.$

Setting $m=2$ and $n=3,$ we get $$f(5)+f(5)=f(2)f(3)+2.$$Then $2(a^5-2a^3+2a^2+a)=(a^2+1)(a^3+2)+2.$ Checking $a=0,$ $a=1,$ and $a=2,$ we find that the only value that works is $a=2.$

Hence, $$f(m+1)=2f(m)-f(m-1)+2.$$The first few values are $$\begin{array}{rl}f(2)& =2f(1)-f(0)+2=5,\\ f(3)& =2f(2)-f(1)+2=10,\\ f(4)& =2f(3)-f(2)+2=17,\end{array}$$and so on. By a straight-forward induction argument, $$f(n)=n^2+1$$for all integers $n.$

We can check that this function works. Therefore, $n=1$ and $s=5,$ so $n\times s=\boxed{5}.$