I am trying to show that $k[x,y,z]/(xz-y^2)\not\cong k[x,y]$. The latter is a UFD, so I am trying to show the former is not. Clearly $x$ is not prime, since $x|xz$ which implies $x|y^2$ but $x|y$. So if I can show that $x$ is irreducible then $k[x,y,z]/(xz-y^2)$ is a not a UFD, because irred implies prime in UFDs.
Attempt: Assume $x$ is reduced into $ab$ in $k[x,y,z]/(xz-y^2)$, then we know there exists a $c$ in $k[x,y,z]$ such that $x=ab+c(xz-y^2)$ in $k[x,y,z]$, or more usefully as $x-ab=(xz-y^2)c$. Someone suggested that I write $a,b$ such they are degree at most 1 in $y$, that is, replace every $y^n$ using $xz$. This means the LHS is degree at most 2 in $y$ and the RHS is degree at least two in $y$ in $x-ab=(xz-y^2)c$. I am not sure where to go from here just by knowing that $c$ has no $y$ terms though. Ultimately I want to show that this forces $x=ab$ in $k[x,y,z]$ which, since $x$ is irred there, shows that either $a,b$ is a unit, which means $x$ is irreducible in $k[x,y,z]/(xz-y^2)$.