= Solution
The statement is true. Fix an isomorphism $\phi:M\otimes_RN\to R^n$, and write the inverse image of the first standard basis vector as
$$
\phi^{-1}(e_1)=\sum_{j=1}^r m_j\otimes n_j.
$$
Define
$$
\psi:N^r\longrightarrow R,
\qquad
(y_j)_j\longmapsto
\operatorname{pr}_1\!\left(\phi\left(\sum_jm_j\otimes y_j\right)\right).
$$
Since $\psi((n_j)_j)=1$, this is a split surjection. Tensor its splitting with $M$. The resulting split surjection has the form
$$
M\otimes_RN^r
\cong(M\otimes_RN)^r
\cong R^{nr} woheadrightarrow M.
$$
Thus $M$ is a direct summand of a finite <free module>, so it is a <projective module>. Symmetry gives the same conclusion for $N$. This is <projectivity of factors of a nonzero finite free tensor product>.
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