Two elements in a non-integral domain which are not associates but generate the same ideal 
Let $\mathbb{K}$ be a field. Let $R$ be the quotient ring $\mathbb{K}[x,y]/(xy^{2})$. Let $\bar{x}$ be the class of $x$ in $R$ (i.o.w. $\bar{x}=x+(xy^{2}))$.
  Prove that $\bar{x}$ and $\bar{x}+\bar{x}\bar{y}$ are not associates but $(\bar{x}) = (\bar{x}+\bar{x}\bar{y})$ (as ideals in $R$).

I proved that the ideals $(\bar{x})$ and $(\bar{x}+\bar{x}\bar{y})$ are the same in $R$: the inclusion $(\bar{x})\subseteq(\bar{x}+\bar{x}\bar{y})$ is given by $\bar{x}=(\bar{x}+\bar{x}\bar{y})-\bar{y}(\bar{x}+\bar{x}\bar{y})$, the other one is given by $\bar{x}\mid(\bar{x}+\bar{x}\bar{y})$.
I don't know how to prove that the two elements are not associates in $R$. By definition, it means that does not exist a unit element $u\in R^{\times}$ such that $\bar{x} = u(\bar{x}+\bar{x}\bar{y})$.
By considering $\bar{x}=(\bar{1}-\bar{y})(\bar{x}+\bar{x}\bar{y})$, I've thought to prove that $\bar{1}-\bar{y}$ is not a unit element in $R$ but I don't know how to  go on with the proof.

I won't use the bar notation. Instead I denote the indeterminates by $X,Y$ and their residue classes by $x,y$.
Suppose that there is an invertible element $u\in R$ such that $ux=x(1+y)$. This means that in $K[X,Y]$ we have $U(X,Y)X-X(1+Y)\in(XY^2)$, that is, $U(X,Y)-(1+Y)\in(Y^2)$ and therefore we can write $U(X,Y)=1+Y+Y^2V(X,Y)$. Since $xy^2=0$ in $R$ we may assume that $V$ is a polynomial only in $Y$, so $u=1+y+y^2v(y)$. But $u$ can't be invertible in $R$ because it lies in a maximal ideal: consider an irreducible polynomial $f\in K[Y]$ such that $f(Y)\mid 1+Y+Y^2V(Y)$, and let $\mathfrak m=(x,f(y))$.
