It is an exercise.
Let $R=k[x_1,\dots,x_n]$ where $k$ is an algebraically closed field. Assuming that
(1) $R$ is Noetherian, and
(2) the maximal ideas of $R$ are precisely the ideals of the form $$ (x_1-a_1,\dots,x_n-a_n) $$ for $a=(a_1,\dots,a_n)\in k^n$
how to prove that $\mathbb{I}(\mathbb{V}(I))=\operatorname{rad}(I)$?
I know the usual prove is to use the so-called The Rabinowitsch Trick, i.e. consider the ideal $(I,1-yf)\subset R[y]$ for some $f\in\mathbb{I}(\mathbb{V}(I))$.
But this requires knowing that $R[y]$ also satisfies $(2)$, which is not given in the question.
So, is there any other ways to proof the strong Nullstellenstaz?
Since the exact question is not quite clear to me, I'll prove that for an arbitrary field $k$ the condition (2) implies that $k$ algebraically closed. Then the Nullstellensatz can be used to answer all the questions posed by the OP.
Consider an irreducible polynomial $f(x_1)\in k[x_1]$ and the ideal $I=\langle f(x_1),x_2,\cdots,x_n\rangle\subset R=k[x_1,\cdots,x_n]$.
Since $R/I=k[x_1]/\langle f(x_1) \rangle$ is a field, the ideal $I$ is maximal and thus by hypothesis of the form $I=\langle x_1-a_1,x_2-a_2,\cdots,x_n-a_n\rangle$.
From this we deduce by intersecting with $k[x_1]\subset R$ :$$ k[x_1]\cap I = \langle f(x_1)\rangle = k[x_1]\cap \langle x_1-a_1,x_2-a_2,\cdots,x_n-a_n\rangle= \langle x_1-a_1\rangle $$ so that $\langle f(x_1)\rangle =\langle x_1-a_1\rangle$ .
Thus any irreducible polynomial $f\in k[x_1]$ satisfies $f(x_1)=c(x_1-a_1)\;(c\in k)$ and we have proved that $k$ is algebraically closed.