No, for MDS codes used in the way it is used in AES there's no other choice, i.e., an MDS code with these dimension must have all entries nonzero in the $M$ matrix.
The MDS matrix $M$ has to be square, mapping 4 bytes to 4 bytes and must have 4 nonzero entries in each row by MDS property.
It also means changing each input byte affects each output byte, helping in diffusion.
Background: by coding theory the code is generated by the $k\times n$ matrix $$G=[I|M]$$
The Matrix converts a length $k$ message $m$ to the codeword $c$ via
where $m,c$ are row vectors. This code has minimum weight $n-k+1$, the maximum possible by singleton bound. Since each row is a codeword the $M$ submatrix must have rows of minimum weight $n-k$. For AES, $n=2k=8$ so $n-k=4.$ So the rows of $M$ must have 4 nonzero entries.