I'm pretty sure this is unknown, though it would be great if I'm wrong.

Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper of Canfield and McKay.

It is also known to be true for $n\le 20$.

To spell out the asymptotics a bit more, the paper cited above shows that $$ M(n,k) = (e^{-1/2}+o(1)) \binom {n}{k}^{\!2n} \bigl( \lambda^\lambda(1-\lambda)^{1-\lambda}\bigr)^{n^2}, \qquad(*)$$ as $n\to\infty$, where $\lambda=k/n$ and $cn/\log n\le k\le n-cn/\log n$ for a particular constant $c$. This expression without the error term is unimodal for $0\le k\le n$. The same formula holds for $k=o(n^{1/2})$ (McKay and Wang, 2003). A new method of Liebenau and Wormald will (I'm 100% confident) show that $(*)$ is also true for the intermediate ranges of $k$, but it is not published yet. Then we will know that $M(n,k)$ is unimodal in $k$ provided $n$ is large enough.

If we just want to know whether $M(n,k)\le M(n,\lfloor n/2\rfloor)$, as asked, and don't care about unimodality, then what remains asymptotically is to show that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le cn/\log n$. Maybe that is not so hard, since the right side is probably more than exponentially larger than the left side.

I'm pretty sure this is unknown, though it would be great if I'm wrong.

Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper of Canfield and McKay.

It is also known to be true for $n\le 20$.

To spell out the asymptotics a bit more, the paper cited above shows that $$ M(n,k) = (e^{-1/2}+o(1)) \binom {n}{k}^{\!2n} \bigl( \lambda^\lambda(1-\lambda)^{1-\lambda}\bigr)^{n^2}, \qquad(*)$$ as $n\to\infty$, where $\lambda=k/n$ and $cn/\log n\le k\le n-cn/\log n$ for a particular constant $c$. The value $c=1/3$ will do. This expression without the error term is unimodal for $0\le k\le n$ and even with the error term it is unimodal if $n$ is large enough. The same formula holds for $k=o(n^{1/2})$ (McKay and Wang, 2003). A new method of Liebenau and Wormald will (I'm 100% confident) show that $(*)$ is also true for the intermediate ranges of $k$, but it is not published yet. Then we will know that $M(n,k)$ is unimodal in $k$ provided $n$ is large enough.

If we just want to know whether $M(n,k)\le M(n,\lfloor n/2\rfloor)$, as asked, and don't care about unimodality, then what remains asymptotically is to show that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$. This follows from the pairing (configuration) model for random $k$-regular bipartite graphs; namely $$ M(n,k) \le \frac{ (nk)! }{ (k!)^{2n} }.\qquad(\#)$$ For large $k$, $(\#)$ is a pretty terrible bound, but if I didn't miscalculate it is sufficient to prove that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le \frac13 n/\log n$.

That completes the proof that $M(n,\lfloor n/2\rfloor)$ is the largest value if $n$ is large enough.

I'm pretty sure this is unknown, though it would be great if I'm wrong.

Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper.

It is also known to be true for $n\le 20$.

I'm pretty sure this is unknown, though it would be great if I'm wrong.

Asymptotically there is a function $A(n)$ such that $M(n,n/2+t)\sim e^{-2t^2}A(n)$ for $t=o(n^{1/2})$, which follows from this paper of Canfield and McKay.

It is also known to be true for $n\le 20$.

To spell out the asymptotics a bit more, the paper cited above shows that $$ M(n,k) = (e^{-1/2}+o(1)) \binom {n}{k}^{\!2n} \bigl( \lambda^\lambda(1-\lambda)^{1-\lambda}\bigr)^{n^2}, \qquad(*)$$ as $n\to\infty$, where $\lambda=k/n$ and $cn/\log n\le k\le n-cn/\log n$ for a particular constant $c$. This expression without the error term is unimodal for $0\le k\le n$. The same formula holds for $k=o(n^{1/2})$ (McKay and Wang, 2003). A new method of Liebenau and Wormald will (I'm 100% confident) show that $(*)$ is also true for the intermediate ranges of $k$, but it is not published yet. Then we will know that $M(n,k)$ is unimodal in $k$ provided $n$ is large enough.

If we just want to know whether $M(n,k)\le M(n,\lfloor n/2\rfloor)$, as asked, and don't care about unimodality, then what remains asymptotically is to show that $M(n,k)\lt M(n,\lfloor n/2\rfloor)$ for $k\le cn/\log n$. Maybe that is not so hard, since the right side is probably more than exponentially larger than the left side.