i just bashed it (it took me not a long time) till i got the ordered pair of $(m,n)$ as $(4,3)$. The difference of the cubes is $64-27=37$, which is prime, and we can show that $37 \equiv \boxed{1}$ mod 6. Im not sure how to solve it without bashing

$m-n|m^3-n^3$ so $m-n=1$ $m^3-n^3=(n+1)^3-n^3=3n^2+3n+1 \equiv 1 \pmod {6}$

$m^3-n^3=(m-n)(m^2+mn+n^2)=p\implies \boxed{m=n+1}\implies m^3-n^3=3n^2+3n+1\equiv1\pmod{6}$

m-n=1, so m^3 -n^3 \equiv 1 \pmod{6}

Since m^3-n^3 is prime, we have that m-n=1 (because m^2+mn+n^2>m-n) so m=n+1. After we substitute m with n+1 we get that m^3-n^3=3n^2+3n+1 which is always congruent to 1 mod 6.