Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.
First, note if a prime $p>3$ divides one of the numbers, it can't divide any other. So it must occur cubed, in the form $p^{3k}$ for some $k$.
Second, note that at least one of the four numbers must be co-prime to both $2$ and $3$; so this number contains only primes $p>3$, and it must therefore be a perfect cube.
And if one of the numbers is a perfect cube, then the product of the other three must also be a perfect cube. So now we have reduced the problem to showing that the product of three out of four consecutive numbers can't be a perfect cube.
There are three cases: $n(n+1)(n+2), n(n+1)(n+3),$ and $n(n+2)(n+3)$. We show by elementary algebra that each of these lies strictly between two consecutive cubes, except for $n(n+1)(n+3)=8$ when $n=1$; so none of them can be a perfect cube, except for this one case. But this case does not lead to a counter-example of the full statement, because the missing number $n+2=3$ is not a perfect cube.
Case 1: $n(n+1)(n+2)$
$n^3<n(n+1)(n+2)=n^3+3n^2+2n<n^3+3n^2+3n+1=(n+1)^3$
Case 2: $n(n+1)(n+3)$
$n(n+1)(n+3)-(n+1)^3=n^2-1$, so unless $n=1$, we have $(n+1)^3<n(n+1)(n+3)<(n+2)^3$.
Case 3: $n(n+2)(n+3)$
$(n+1)^3=n^3+3n^2+3n+1\underset{3n<2n^2+5}{<}n^3+5n^2+6=n(n+2)(n+3)\underset{n(n+3)<(n+2)^2}{<}(n+2)^3$
Best Answer
Suppose that $n$ is an integer such that $n = a^2 = b^3$ for some integers $a, b$.
Let $p$ be any prime.
How many times does $p$ divide $n$?
Can you show that it must be a multiple of $6$?
Hence conclude that $n$ must be of the form $c^6$, i.e. a perfect sixth power.