Run AES for 2 rounds
The difference of one byte before MixColumn will propagate to the whole column. The goal of this step is to mix the four bytes together to get another four bytes. Look at the definition and try it yourself.

Also, there is some nice property here. There are $256$ possible values for the XOR difference of the two bytes and since MixColumn is linear, the XOR difference of the two columns after applying MixColumn will be the same as applying MixColumn to the difference of the columns before. So there are only $256$ possible values for the entire difference of the two columns after MixColumn.

To make it clear: if $C_1$ and $C_2$ are the two columns that differ in only one of the four bytes, then there are $256$ values for $C_1 \oplus C_2$, and we have $\mathrm{MC}(C_1 \oplus C_2) = \mathrm{MC}(C_1) \oplus \mathrm{MC}(C_2)$, so only $256$ possible values for $\mathrm{MC}(C_1) \oplus \mathrm{MC}(C_2)$.

I think it's a nice property.