Advantage of $\operatorname{GF}(2^8)$ over $\mathbb Z/2^8\mathbb Z$ in AES/Rijndael
One important property of the mixColumns step is that it is Maximum Distance Separable (MDS).  That is, if $M$ is our multiplication matrix, if you take any two distinct input vectors $V$ and $V'$, and compute $M \cdot V$ and $M \cdot V'$, the total number of bytes distinct in $V$ and $V'$ plus the total number of bytes distinct in $M \cdot V$ and $M \cdot V'$ will always be at least 5 (the dimension of the matrix $M$ plus 1).

It turns out that this is an important property used in the proof of security of AES against differential and linear cryptanalysis.

Now, if you replace the $GF(2^8)$ multiplications within the MDS matrix with $Z/256$ multiplications, it turns out that you cannot have the MDS property; neither the matrix used within AES, nor any other matrix has the MDS property.

This is fairly easy to show; to start with, consider $V = \{0,0,0,0\}$ and $V'=\{128,0,0,0\}$; these obviously differ in 1 bytes, and so for the MDS property to hold, all for values of $M \cdot V$ and $M \cdot V'$ must differ.  However, if lsbit of $M[0,0]$ is 0, then the first byte of $M \cdot V$ and $M \cdot V'$ will be the same, and hence MDS will not hold, and hence the lsbit of $M[0,0]$ must be a 1.  Along the same line, if we consider the other bytes of $M \cdot V$ and $M \cdot V'$, and additionally, $V'=\{0,128,0,0\}, \{0,0,128,0\}, \{0,0,0,128\}$, we find that all bytes of $M$ must have an lsbit of 1.  Finally, we consider $V'=\{128,128,128,128\}$.  If all the lsbits of $M$ are 1, then $M \cdot V = M \cdot V'$; that is also a violation of the MDS property (as well as making decryption difficult...)