Ice cream quiz

This problem was posted in an ice cream shop in Oxford, England. I found it surprisingly difficult.

Here is my solution. Suppose that $x + 3 = a^3$ and $x^2 + 3 = b^3$. After substituting,

$a^6 > (a^3 - 3)^2 + 3 = b^3$

which implies that $a^2 - 1 \ge b$. Therefore

$(a^3 - 3)^2 + 3 \le (a^2 - 1)^3$

$a^6 - 6a^3 + 12 \le a^6 - 3a^4 + 3a^2 - 1$

$3a^4 - 6a^3 - 3a^2 \le -13$

$3a^2 (a^2 - 2a - 1) \le -13$

But the left side is positive for $a \ge 3$, so there are no whole number solutions.

Image credit: Janet McKnight

Thanks to @samuelprime and @nodrogadog for pointing out mistakes in earlier versions of this post.