Let $\phi: R \to R'$ be a ring map, and $M$ is an $R$-module. The ring map induces map of specs $\Phi : \operatorname{Spec}(R') \to \operatorname{Spec}(R)$. Then
$$
\operatorname{Supp} (M \otimes_R R') \subset {\Phi}^{-1}(\operatorname{Supp}(M)),
$$
with equality for finitely generated modules.
I have a "proof" that gives equality for all modules and I can't find my mistake.
Choose a set $\{m_i\}_{i \in I}$ of generators of $M$. Module $M$ is a sum of cyclic modules $M = \sum_i R m_i$, hence
$$
\operatorname{Supp}(M) = \cup_i \operatorname{Supp}(R m_i).
$$
For cyclic modules $R m_i = R/J_i$, where $J_i = \operatorname{Ann}(m_i)$, so $\operatorname{Supp}(R m_i) = V(J_i)$.
For right hand side we get
$$
{\Phi}^{-1}(\operatorname{Supp}(M))= \Phi^{-1}(\cup_i V(J_i))=\cup_i \Phi^{-1}(V(J_i))=\cup_i V(J_i R').
$$
For the left hand side we use similar argument: $M \otimes_R R' = \sum_i R'(m_i \otimes 1)$, so
$$
\operatorname{Supp}(M \otimes_R R') = \cup_i \operatorname{Supp}(R'(m_i \otimes 1)).
$$
As before $\operatorname{Supp}(R'(m_i \otimes 1))=V(K_i)$, where $K_i = \operatorname{Ann}(m_i \otimes 1)=\operatorname{Ann}(m_i)R'=J_i R'$, so
$$
\operatorname{Supp}(M \otimes_R R') = \cup_i V(J_i R').
$$
I did not use that $M$ is finitely generated.
Best Answer
Check out the relevant lemma from the stacks project
https://stacks.math.columbia.edu/tag/07T8
The point is that you cannot compute the annihilator of the element $m\otimes1$ as you did unless you know that $\phi\colon R\to R’$ is flat.