Since a is even, a's prime factorization must contain at least one factor of 2. Suppose a is not a multiple of 4, so a has only one factor of 2. Thus a^2 has exactly two factors of 2. However, the equality requires a^2 to be expressible as b^3 with b an integer. This would require a^2 to have all entries in the prime decomposition to be to the power of a multiple of three. The number of factors of 2 in a's prime decomposition is two, which is not a multiple of 3. Thus our supposition that a could fail to be a multiple of 4 is false. QED.

There is an implied other side to this: showing that there is at least one case where the equality holds. I put forward 8^2 = 4^3 = 64.

But more can be said! In fact, if a^2 = b^3, then a is a cube and b is a square (unless they're both 1 or 0). But that is for another node... except for the implication that under these very same conditions, we know that a is a multiple of 8!