Here is an idea and a proof sketch.
Solution: (c*a, c*b) for cubic numbers a and b and integer c >= 1. It is
easy to see those are solutions, but takes a proof to show that they are the
only solutions.
Observation: if (a,b) is a solution, then (k*a, k*b) is also a solution, for
integer k >= 1.
Claim: If (a, b) is a solution, then both a and b must be cubic numbers.
Now let's prove that claim. W.l.o.g. we may assume gcd(a,b) = 1, and we may
also assume that a is not a cubic number. Let p be a prime factor of a. We
know that b does not divide p.
Let u = a^1/3 and v = b^1/3, also let a = p*q
(u + v)^3 = a + b + u^2*v + u*v^2
Notice that u^2*v * u*v^2 = u^3 * v^3 = a * b is an integer, so u^2*v is an
integer iff u*v^2 is an integer.
u^2*v = p^2/3 * q^2/3 * b^1/3
This cannot be an integer because p is a prime and u^2*v must divide p if u^
2*v is an integer. (* This step is hand wavy and some number theory expert
please tell me what theorem needed here. *)
For the same reason u*v^2 is not an integer.
Q.E.D.