Here are some thoughts, too long for a comment. We can assume that $c$ is not a rational square, otherwise we have many solutions with $x=y$ and $z\neq 0$.
We have $x^2-1=c_1 z^2$ and $y^2-1=c_2 z^2$ with $c_1 c_2=c$. Hence $(x,y,z)$ lies on the intersection on of two space quadrics. As $c_1\neq c_2$, the intersection is an elliptic curve isomorphic to the plane curve $$ Y^2=-8(X-1)(X+1)(-c_1X-2c_2+c_1),$$ see Pinch: Square values of quadratic polynomials here for details. Namely, apply Proposition 1.1 there with $(a,b,c,d,e,f)=(1,-c_1,1,1,-c_2,1)$ and $(p,q,r,s)=(1,0,1,1)$. With the notation $2U=c_1(X-1)$ and $8V=c_1Y$ the last equation becomes $$ V^2=U(U+c_1)(U+c_2). $$ A solution $z\neq 0$ corresponds to $V\neq 0$, hence the problem boils down to whether the last equation can be solved with $V\neq 0$ for some decomposition $c=c_1c_2$.