Firstly, note that the equation $x^2 - py^2 = 1$ has infinitely many solutions in positive integers. (Pell's equation)
Then for every prime number $p$ there exist infinitely many positive integers $s$ and $t$ such that $s^2 - 1 = pt^2$.
Putting $x = s^2 - 1$, $y = s + 1$, $z = s - 1$ we have
$$(x^2 + s^2 - 1)(y^2 + s^2 - 1)(z^2 + s^2 - 1) = \\ = (s^2 - 1)s^2(s + 1)(2s)(s - 1)(2s) = ((s^2 - 1)2s^2)^2.$$
It is remained to check that $x, y, z \ne t$.
Note that $x = s^2 - 1 = pt^2 \ne t$.
If $y = s + 1 = t$ then
$$(s-1)(s+1) = pt^2 = p(s+1)^2 \Rightarrow s-1 = p(s+1) > s+1 > s-1.$$
If $z = s - 1 = t$, then
$$(s-1)(s+1) = pt^2 = p(s-1)^2 \Rightarrow s+1 = p(s-1) \ge 2(s-1) > s+1,$$
for $s > 3$.