Why is $F[x,y,z]/(x^2-yz)$ a domain? How can I show that the ring $R=F[x,y,z]/(x^2-yz)$ is an integral domain? (Here $F$ is a field.)

I tried to prove this by contradiction. Suppose there are $\bar f, \bar g \in R-\{0\}$ such that $\bar f\bar g =0 $ in $R$, i.e. $fg$ is divisible by $x^2-yz$ in $F[x,y,z]$. But I don't know how to proceed.
Consider $p(x,y,z):=x^2-yz$ as a polynomial in $R[x]$, where $R:=F[y,z]$ is a unique factorization domain.  If we can show that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$, then we are done.  First, note that $y$ is a prime element of $R$.  The constant term of $p(x,y,z)$ as a polynomial in $x$ is $-yz$ which is not in the ideal generated by $y^2$ of $R$.  The coefficient of $x$ in $p(x,y,z)$ is $0$, which belongs in the ideal generated by $y$ of $R$.  Using Eisenstein's Criterion for integral domains,  we conclude that $p(x,y,z)$ is irreducible in $R[x]=F[x,y,z]$.  Since $F[x,y,z]$ is a unique factorization domain, $p(x,y,z)=x^2-yz$ is a prime element.  That is,
$$F[x,y,z]/\langle x^2-yz\rangle$$
is an integral domain.
$x^2-yz$ is irreducible, so $x^2-yz$ is prime, so $F[x,y,z]/(x^2-yz)$ is integral domain.