Let $I = (x^2, y)$ be an ideal of $\mathbb{Q}[x,y]$ show that Rad$(I)$ = $(x,y)$ and $I$ is a primary ideal that is not a power of a prime ideal

Question

Let $I = (x^2, y)$ be an ideal of $\mathbb{Q}[x,y]$ show that Rad$(I)$ = $(x,y)$ and $I$ is a primary ideal that is not a power of a prime ideal.

I can see that $(x,y) \subset Rad(I)$ b/c $x^1, y^2 \in (x,y^2)$

Further $x \in (x,y)$ and $y^2 \in (x,y)$ so this clearly gives $(x,y^2) \subset (x,y)$ and $(x,y)$ is maximal thus $Rad(I) \subset (x,y)$ hence they are equal. For $I$ is primary: since $Rad(I) = (x,y)$, Rad$(I)$ is maximal

I think the correct course is $I$ is primary if every zero-divisor in $\mathbb{Q}[x,y]/I$ is nilpotent