Note that $253^2=64009$ and $40^3=64000$ so the difference $9$ is possible. Let us show that nothing smaller works. First of all, it is clear that the difference is always odd and divisible by $3$, so the only possible smaller value is $3$. But if $253^m-40^n=3$, then mod $7$ we get $n \equiv 1 \pmod{6}$ while mod $9$ we get $n \equiv 2 \pmod{3}$. On the other hand, if $253^m-40^n=-3$, then mod $7$ we get $n \equiv 2 \pmod{6}$ while mod $9$ we get $n \equiv 1 \pmod{3}$. In any case, we get a contradiction, hence $9$ is indeed the smallest possible difference.