Given $z^2-1\mid x^2z^2-1$, prove $\frac{x^2z^2-1}{z^2-1}$ is never prime, for $x$, $z$ integers such that $x>z>1$. 
Given $z^2-1\mid x^2z^2-1$, prove that $\frac{x^2z^2-1}{z^2-1}$ can never be prime, assuming $x$, $z$ are integers such that $x>z>1$.

So far I have tried taking mod a lot of different numbers, but I cannot find solution. I also tried writing as a quadratic and using quadratic formula, but that doesn't work either. Please help.

Answered for my own benefit, as I already had seen the above two answers first******
Note that $x^2z^2-1 = (xz-1)(xz+1)$. Then, as $\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1}$ is an integer, it follows that $z^2-1$ can be written as follows: $z^2-1 = ab$ for some positive integers $a$ and $b$, that satisfy the following:

*

*$a|(xz-1)$ say $xz-1=ca$ and


*$b|(xz+1)$ say $xz+1=bd$.
[One of $a,b$ may be $1$]. But then, as $xz-1 > z^2-1$ [because $x>z$], it follows that $c$ as in 1. above must be an integer greater than $1$, and similarly, $d$ as in 2. above must be an integer greater than $1$. But then that gives:
$$\frac{x^2z^2-1}{z^2-1} = \frac{(xz-1)(xz+1)}{z^2-1} = \frac{(ca)(db)}{ab} = cd,$$ where $c$ and $d$ are both integers greater than $1$.
Thus, as $cd$ is clearly compositive for such $c$ and $d$, it follows that $\frac{x^2z^2-1}{z^2-1}=cd$ cannot be prime.
In general, the following is true:

Let $A,B,C$ be positive integers such that both $\frac{AB}{C}$ is integral, and $C<\min\{A,B\}$. Then $\frac{AB}{C}$ is always not prime.

