$R = \mathbb{Q}[x,y,z]$ and let bars denote passage to $\mathbb{Q}[x,y,z]/(xy-z^2)$. Prove that $\bar{P} = (\bar{x},\bar{z} )$ is a prime ideal. Show that $\bar{xy} \in \bar{P}^{2}$, but no power of $\bar{y}$ lies in $\bar{P}^2$.
Let $P = (x,z)$ it's clear that $xy \in (x,z)$ and $z^2 \in (x,z)$ thus $ xy-z^2 \in (x,z)$ further $\overline{(xy-z^2)} \subseteq (\bar{x},\bar{y})$
so then $(\bar{x},\bar{z})/\overline{(xy-z^2)}$ is an ideal of $\mathbb{Q}[x,y,z]/(xy-z^2)$. Further, by Isomorphism Theorem 3
$\displaystyle \frac{\mathbb{Q}[x,y,z]/(xy-z^2)}{(\bar{x},\bar{z})/(xy-z^2)} \simeq \displaystyle\frac{\mathbb{Q}[x,y,z]}{(\bar{x},\bar{z})} \simeq \mathbb{Q}[y] $ which is an integral domain thus $\bar{P} = (\bar{x},\bar{z})$ is prime
Now, $\bar{P}^2 = (\bar{x}^2, \bar{xz}, \bar{z}^2)$ Also $\overline{xy-z^2} = \overline{0}$ so $\overline{xy} = \overline{z^2}$
so $\overline{xy} \in \overline{P}^2$
Finally, $\overline{P}^2 \subseteq \overline{P}$, suppose $\overline{y}^n \in \overline{P}^2$ then $\overline{y}^n \in \overline{P}$ then $y^n \in P$ so $0 = y^n \in \mathbb{Q}[y]$
but $\mathbb{Q}[y]$ is an integral domain and the only nilpotent element is $0$ hence no powers of $\overline{y}$ are in $\overline{P}^2$
Not sure this is correct