For a right $R$-module $M$, let $M^*$ denote the Pontrjagin dual of $M$, defined as $$M^* := \operatorname{Hom}_\mathbb{Z}(M, \mathbb{Q}/\mathbb{Z})$$
We call two right $R$-modules $M_1, M_2$ flat equivalent if there exist flat modules $F_1, F_2$ such that $(M_1 \oplus F_2)^* \cong (M_2 \oplus F_1)^*$.
It's straightforward to show that if $M_1$ and $M_2$ are flat equivalent, then $\operatorname{Tor}_n(M_1, N)^* \cong \operatorname{Tor}_n(M_2, N)^*$ for $n>0$.
(The argument could go: WLOG consider flat resolutions of $M_1 \oplus F_2$ and $M_2 \oplus M_1$ since $\text{Tor}_n$ is unaffected by the $F_i$ summands for $n>0$ (ii) applying $^*$ to the resolutions yields injective resolutions and converts $\text{Tor}_n(M_1 \oplus F_2, N)$ into $\text{Ext}^n_\mathbb{Z}(N, (M_1 \oplus F_2)^*)$ and similarly for $M_2 \oplus F_1$, i.e. the relevant Tor groups are converted into cohomologies of injective resolutions of isomorphic modules.)
One nice consequence of this is the fact that $\operatorname{Tor}_n(M_1, N) = 0$ iff $Tor_n(M_2, N) = 0$ for $n>0$.
A much stronger statement is true, writes T.Y. Lam$^1$.
$\text{Tor}_n(M_1, N) \cong \operatorname{Tor}_n(M_2, N)$ when $M_1$ and $M_2$ are flat equivalent
I'd really like to understand this. Could anyone provide a reference or a sketch of proof?
$^1$ T.Y. Lam, Lectures on modules and rings, last paragraph of $5D$ (p. 187 in first edition)