 Strict descent of powers of a nonzero proper ideal in a Noetherian domain

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.