Prime-free sequence 
Prove that there exist infinitely many positive integers $k$ such that
  the sequence $\{x_n\}$ satisfying
$$ x_1=1, x_2=k+2, x_{n+2}-(k+1)x_{n+1}+x_n=0(n \ge 0)$$ does not
  contain any prime number.

I found a similar question: The sequence ${x_n}$ satisfies $x_{n+2}=(k+1)x_{n+1}-x_n, x_0=1, x_1=k+2$ and every term of the sequence is either $1$ or composite. Prove that the set of such $k$ is infinite.
Solution: Let $\alpha=\frac{\sqrt{k+3}+\sqrt{k-1}}{2}$, $\beta=\frac{\sqrt{k+3}-\sqrt{k-1}}{2}$, then$$x_n=\frac{\alpha^{2n+1}-\beta^{2n+1}}{\alpha-\beta}.$$We consider$$x_{n}^2=\frac{(\alpha^{2n+1}+\beta^{2n+1})^2-4}{k-1},$$let $a_n=\alpha^{2n+1}+\beta^{2n+1}$, $a_0=\sqrt{k+3}$, $a_1=\frac{(k+3)^{\frac{3}{2}}+3\sqrt{k+3}(k-1)}{4}$. We let $k+3=u^2$, and $2|u$, then $a_n$ is a positive integer, we have$$(k-1)x_{n}^2=(a_{n}+2)(a_{n}-2).$$If $x_n=p$ is a prime, $p$ is lagre enough when $n$ is lagre enough, $gcd(a_n-2, a_n+2)=gcd(a_n-2, 4)\leq 4$, it follows that $p^2\mid a_n+2$, or $p^2\mid a_n-2$, a contradiction.
Is a different solution like this?

I tried using the Z transform. From the question, we conclude that $x_0 = -1$. And taking the Z transform we get:
$$ x_{n+2} - (k+1)x_{n+1}+x_n = 0 $$
$$ z^2X(z)-x_0z^2-x_1z-(k+1)(zX(z)-x_0z) + X(z) = 0 $$
$$ X(z) = \frac{(k+1)z+z-z^2}{z^2-(k+1)z+1} $$
if we let $a=k+1$ then after taking the inverse  Z transform we get:
$$  x_n = 2^{-n-1}\frac{ (-\sqrt{a^2-4}+a+2)(\sqrt{a^2-4}+a)^n - (\sqrt{a^2-4}+a+2)(a-\sqrt{a^2-4})^n }{\sqrt{a^2-4}}$$
This might be helpful for further solving.
For $a=2$ which is $k=1$ the sequence becomes:
$$ x_n = 2n - 1 $$ and it contains infinitely many prime numbers since it is a sequence of odd numbers and we have a contradiction.
