My solution:$x=0:b=n$ ,$x=1:c=m+n+1$, $x=n:a=n^2+mn+n$ Then $a=bc$

Maybe when $n=0$, $n=1$ is need some work for the above attempt. However, let us consider more general question of whether there exist infinitely many. The answer is in the affirmative. Case I: $m$ is even. Then, we just have $A(m, n) = \{x^2+k\mid x\in \mathbb{Z}\}$ where $k=n-m^2/4$. Now, we can rewrite $x^2+k=(y^2+k)(z^2+k)$ as $x^2+k = (yz+k)^2+k(y-z)^2$. Thus, we can just take $(x,y,z)=(t^2+t+k, t+1, t)$. Case II: $m$ is odd. Then, we just have $A(m, n) = \{\frac{(2m+1)^2+k}{4} \mid m\in \mathbb{Z}\}$ where $k=4n-m^2$. Now, we can rewrite $4(x^2+k)=(y^2+k)(z^2+k)$ as $4x^2+4k=(yz+k)^2+k(y-z)^2$. Thus, we can just take $(x,y,z)=\left(\frac{t^2+2t+k}{2}, t+2, t\right)$ for $t$ odd.