$R[x]/(x^2)\simeq R[x]/((x-1)^2)$

Question

How do I prove that $R[x]/(x^2)\simeq R[x]/((x-1)^2)$? I tried to use the universal property of quotient rings as follows. Consider the composition $$R[x]\rightarrow R[x]\rightarrow R[x]/((x-1)^2)$$ given by $x\mapsto x-1\mapsto x-1+(x-1)^2$. The first map is an isomorphism. The kernel of the second map is $(x-1)^2$. The kernel of the composition is therefore $(x-1)^2$. If an ideal $I$ in $R[x]$ lies in the kernel, then the composition factors through $R[x]/I$. But the ideal $(x^2)$ doesn't lie in the kernel, so I cannot apply this universal property. Is is the wrong track?