How to check that AES Matrix is MDS? I want to understand how can I prove that M matrix in AES is MDS. I know that a matrix is MDS if every determinant of every square submatrix is different from 0. 

I don't get this. How much submatrix does it have? If I swap rows or columns, does that change something? Do I have to calculate the determinant in a certain mode?
My answer focuses on the AES matrix only. In general, an $A$ matrix with all submatrices having full rank generates an MDS code when concatenated by an identity matrix of the right size to form $[I|A]$, but that's really a coding theory matter.

As to the specific question, this is basic linear algebra but you need to work over the field that AES is defined over. Let the $k\times k$ matrix (like the AES mixing matrix) be
$$A=\left[
\begin{array}{cccc}
a_{1,1}&amp;a_{1,2}&amp;\cdots&amp;a_{1,k}\\
 \vdots &amp;&amp;&amp; \vdots \\ 
a_{k,1}&amp;a_{k,2}&amp;\cdots&amp;a_{k,k}
\end{array}
\right]
$$

The $1\times1$ submatrices are all nonzero, they're just the entries of $A.$

The $2\times 2$ submatrices: choose any two rows or columns. you can also use the fact that the matrix is circulant to reduce the cases.

So the matrix
$$\left[\begin{array}{cc}
a_{1,1}&amp;a_{1,3}\\ 
a_{2,1}&amp;a_{2,3}
\end{array}
\right]
$$

is the same matrix as
$$\left[\begin{array}{cc}
a_{2,2}&amp;a_{2,4}\\ 
a_{3,2}&amp;a_{3,4}
\end{array}
\right]
$$
and thus has the same determinant, etc.

The $4\times$ submatrix is the whole matrix. 

Row and column swaps only change the sign of the determinant and can be ignored.

Edit: You need to use the finite field with 256 elements as explained in the link below.

How to use the Extended Euclidean algorithm to invert a finite field element?

This is NOT mod 256 arithmetic.