On the system of Diophantine equations (m2−1)r+b2=c2 and (m2−1)x+by=cz

Given positive integers r and m, one can create a positive integer solution (b,c) to the first equation in the title by setting b and c as 2b=(m+1)r−(m−1)r and 2c=(m+1)r+(m−1)r. In this note we show that there are only finitely many pairs (r,m) with r≡2(mod4) and m even such that the second equation in the title holds for some triple (x,y,z) of positive integers with (x,y,z)≠(r,2,2).