In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of side length $L$. Then $\lambda_2(M_2)=\pi^2/L^2$ (possible eigenfunction $u=\cos \pi x/L$). If we now take $M_1$ as a thin set close to the diagonal, then we obtain approximately the one-dimensional Neumann eigenvalue of an interval of that length: $\lambda_2(M_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda_2(M_2)$.

In view of Terry's comment on Michael's answer, it's perhaps worth pointing out that this monotonicity also fails if both domains are required to be convex. We can take $M_2=[0,L]^2$ as a square of side length $L$. Then $\lambda_2(M_2)=\pi^2/L^2$ (possible eigenfunction $u=\cos \pi x/L$). If we now take $M_1\subseteq M_2$ as a thin rectangle close to the diagonal, then we obtain approximately the one-dimensional Neumann eigenvalue of an interval of that length: $$ \lambda_2(M_1)\simeq \pi^2/(\sqrt{2}L)^2<\lambda_2(M_2) $$