Here is a very rough sketch, details to be filled in. Let $p_1=p$, $p_2=q$. Note that, $p_3=p+q+k,p_4=p+2q+2k$, and continuing in this manner, we deduce that, $p_n=F_{n-2}p+F_{n-1}q+(F_n-1)k$, where $F_n$ is the $n^{th}$ Fibonacci number. Simply observe that, $F_1=F_2=1$, and going backward, $F_0=0$. Since there are $p^2$ possibilities for the pair $(F_n,F_{n+1})$ modulo $p$, it should repeat itself at some point, that is to say, there is a sequence $(n_k)_{k=1}^\infty$ along which, $F_{n_k-1}\equiv 0\pmod{p}$, and $F_{n_k}\equiv 1\pmod{p}$. Thus, $p_{n_k}\equiv 0\pmod{p}$, for every $k\in\mathbb{N}$, which yields that for $k$ sufficiently large, $p_{n_k}$ is never a prime.

Note if we set the solution $p_n = m^n$, the equation becomes $m^2=m+1$ So the solution to homogeneous part: $p_n=A(\frac{1+\sqrt5}{2})^n+B(\frac{1-\sqrt5}{2})^n$ A typical solution of the relation is $p_n=-k$ for all $k$. By adding the two parts together, we obtain $p_n=A(\frac{1+\sqrt5}{2})^n+B(\frac{1-\sqrt5}{2})^n-k$. If we take the whole recursion mod $p_1$, we will get a periodic sequence, meaning it there will be another number also $\equiv 0\pmod{p_1}$. If $A\geq0$, the sequence eventually increases, and the sequence will not fully be a prime sequence. By the same token, $A$ cannot $\leq0$. $A=0$. What remains is $p_n =B(\frac{1-\sqrt5}{2})^n-k$, but since all prime numbers $p_n \in \mathbb{Z}$ , so $B$ also $=0$. $p_n=-k \Rightarrow p_n$ remains constant for all $p_n$, $k=-p$ for any $p$.