Is "two or zero" matching in a bipartite graph NP complete?
The answer here seems to imply there is a more general result. For this particular case, here is a self contained way to reduce the problem to maximum weight perfect matching. Assume $k$ is even.

Given $G=(L\cup R, E)$, we construct a new graph $G'=(V',E')$ as follows, let $|R|=n$.


Add vertices in $R$ to $V'$.
For each vertex $v \in L$, add vertices $v_1,v_2$ to $V'$.
Add vertices $t_1,\ldots,t_{n-k}$ to $V'$.
Add edge $v_1v_2$ to $E'$ for each $v\in L$. It has weight $0$.
Add edge $v_iu$ to $E'$ for each $vu\in E$,$u\in R,v\in L,i\in \{1,2\}$. It has weight $1$.
Add edge $ut_i$ for each $u\in R$ and $1\leq i\leq n-k$. It has weight $0$.


Claim: The maximum weight perfect matching in $G'$ has value $k$ iff the "two or zero" matching has a solution of at least $k$ edges.