Are there integers nonzero integers $a,b,c,d,x,y,m,n,p$ such that $(a^2-mb^2)(c^2-nd^2)=x^2-py^2$? ($m,n,p)$ are square free non equal integers.

Question

We are all familiar with Fibonacci-Brahmagupta's identity:

$$(a^2-mb^2)(c^2-md^2)=(ac+ mbd)^2-m(ad+bc)^2$$ I am trying to find whether there is a similar identity: $$(a^2-mb^2)(c^2-nd^2)=x^2-py^2$$ where $p,m,n$ are not all equal.

If this problem has already been solved, please save me the trouble by pointing me at a reference. Otherwise, any hint will be greatly appreciated.Thanks.