No reference, but the proof is trivial. If $ab = 0$, then $x^{2n} | ab$ (for any $n$), so since $x$ is prime $x^n | a$ or $x^n | b$, and since this is true for all $n$, either $a$ or $b$ lies in $\bigcap_n (x^n)$ and hence equals $0$. There's no need to assume $R$ is Noetherian.

No reference, but the proof is trivial. If $ab = 0$, then $x^{2n} | ab$ (for any $n$), so since $x$ is prime and a nonzerodivisor, $x^n | a$ or $x^n | b$, and since this is true for all $n$, either $a$ or $b$ lies in $\bigcap_n (x^n)$ and hence equals $0$. There's no need to assume $R$ is Noetherian.