(1) Use proof by contradiction, suppose $z$ is even, that is $z = 2z_1$. The original equation can be rewritten as: $z_1(2xz_1+1)^2 = (5z_1+y)(4z_1+y)$.
Let $w = (y, z_1)$, so let $y = wy_0, z_1 = wz_0$, where $(y_0, z_0) = 1$.
$$\Rightarrow z_0(2xwz_0+1)^2 = w(5z_0+y_0)(4z_0+y_0)$$
Because $(z_0, 5z_0+y_0) = (z_0, 4z_0+y_0) = 1$ and $(w, (2xwz_0+1)^2) = 1 \Rightarrow w|z_0, z_0|w$
So $w = z_0$ and $(2xwz_0+1)^2 = (5z_0+y_0)(4z_0+y_0)$
Because $(5z_0+y_0, 4z_0+y_0) = (z_0, 4z_0+y_0) = (z_0, y_0) = 1$
Therefore we may set $5z_0+y_0 = m^2, 4z_0+y_0 = n^2$, where $m, n$ are positive integers, so $m > n$, that is $m-n \ge 1$.
$$\Rightarrow w = z_0 = m^2-n^2 \Rightarrow 2xw^2+1 = 2xwz_0+1 = mn,$$
Therefore $mn = 1+2xw^2 = 1+2x(m^2-n^2)^2 = 1+2x(m-n)^2(m+n)^2 \ge 1+2x(m+n)^2 \ge 1+8xmn \ge 1+8mn$
That is $7mn \le -1$ (contradiction), so $z$ is not even.

(2) Let $w = (y, z)$ and $y = wy_0, z = wz_0$, where $(y_0, z_0) = 1$.
$$\Rightarrow z_0(xwz_0+1)^2 = w(5z_0+2y_0)(2z_0+y_0)$$
Because $(y_0, z_0) = 1$ and $z_0$ is odd,
so $(z_0, 2z_0+y_0) = 1 = (z_0, 5z_0+2y_0) \Rightarrow z_0|w$,
and also $(w, (xwz_0+1)^2) = 1 \Rightarrow w|z_0$
So $z_0 = w, z = wz_0 = w^2$, which completes the proof.