Injectivity of $I\leadsto V_I(-)$ and relation to Hilbert's Nullstellensatz

Question: "My second question is: does Hilbert's Nullstellensatz link in any way with these observations? They remind me of the bijection between radical ideals and algebraic varieties, but I can't see clearly the connection, if it exists. Thanks for any clarification."

Answer: It seems your claims follow from the nullstellensatz in the following way: If $k \subseteq K$ is the algebraic closure of a field $k$ and if $I,J \subseteq R:=k[x_1,..,x_n]$ let

$$V(I):=\{(a_i)\in K^n: f(a_i)=0\text{ for all $f\in I$}\}.$$

It follows by the HNS that $I(V(J)) = \sqrt{J}$ is the radical of $J$. Hence if $I,J$ are radical ideals with $V(I)=V(J)$ it follows $I=J$. Hence if there is an equality of functors $V_I(-)=V_J(-)$ it follows $V(I)=V(J)$ and hence $I=\sqrt{I}=\sqrt{J}=J$ and hence the correspondence $I \leadsto V_I(-)$ is injective on ideals.

Question: "My first question is: if I only consider radical ideals, then the association $I\leadsto V_I(-)$ returns injective? I would say yes because if $I=\sqrt{I}$, then $k[x_1,\dots ,x_n]/I$ is reduced and so $V_I(-)$ is representable also as a functor on $\mathbf{Alg}^{\mathbf{red}}_k$."

Answer: A similar argument proves this claim. If there is an equality of functors $V_I(-)=V_J(-)$, it follows $V_I(K)=V_J(K)$ hence $V(I)=V(J)$ and hence $I=\sqrt{I}=\sqrt{J}=J$.