Let $n \in \mathbb{Q}$ a number that respects: $\sqrt{n}+\sqrt[3]{n} \in \mathbb{N}$. What can we say about $n$?
My answer is that $n$ must be a perfect square, a perfect cube and an integer at the same time. The problem is showing that $n=\frac{r^6}{q^6}$, because I think that I did it, but using a long way, so I'd like to know if there's a shorter way without many calculations.
If $\sqrt{n} + \sqrt[3]{n} = m$, then $\sqrt{n} = m - \sqrt[3]{n}$. Squaring both sides yields
$$\begin{equation} \tag{$\ast$} \sqrt[3]{n}^2 - 2m \sqrt[3]{n} + m^2 - n = 0. \end{equation}$$
Now if $n$ is not a cube of a rational number, the polynomial $x^3 - n$ does not have any rational roots, so it is irreducible over $\mathbb{Q}$ (since any factorization would involve a factor of degree one). Thus $\left[ \mathbb{Q} \big[ \sqrt[3]{n} \big] : \mathbb{Q} \right] = 3$, so $(\ast)$ may not hold. Hence $n$ is a cube of a rational number.
Now $\sqrt{n} = m - \sqrt[3]{n}$ is rational, so $n$ is also a square of a rational number, therefore $n = a^6$ for some $a \in \mathbb{Q}$.
Edit: since my solution involves some field theory the reader may be unfamiliar with, I will introduce the necessary part below.
Let $\mathbb{Q}[x]$ stand for the set of all polynomials with rational coeffiecients.
Claim. Let $n \in \mathbb{Q}$ and $\alpha = \sqrt[3]{n}$. Suppose the polynomial $p(x) = x^3 - n$ is irreducible over $\mathbb{Q}$, that is, whenever $p(x) \equiv u(x) \cdot v(x)$ for some $u(x), v(x) \in \mathbb{Q}[x]$, either $u(x)$ or $v(x)$ is constant. Then there is no nonzero polynomial $q(x) \in \mathbb{Q}[x]$ of degree at most $2$ such that $q(\alpha) = 0$.
Proof. Suppose that $q(\alpha) = 0$ fo some nonzero polynomial $q(x) \in \mathbb{Q}[x]$ of degree at most $2$. By polynomial division we find $u(x), r(x) \in \mathbb{Q}[x]$ such that $r(x) \equiv 0$ or $\deg r(x) < 2$ and
$$p(x) \equiv u(x) \cdot q(x) + r(x).$$
Substituting $x = \alpha$, we get
$$0 = u(\alpha) \cdot 0 + r(\alpha),$$
i.e. $r(\alpha) = 0$.
If $r(x) \equiv 0$, we get a contradiction, because $p(x) \equiv u(x) \cdot q(x)$ is a factorization of $p(x)$ over $\mathbb{Q}$ with non-constant factors.
If $\deg r(x) = 0$ so $r(x)$ is constant nonzero, this is an immediate contradiction.
If $\deg r(x) = 1$, then $r(\alpha) = 0$ implies $\alpha \in \mathbb{Q}$ so $p(x)$ may be reduced by the factor $x-\alpha$, which is a contradiction.
So the claim is proved. $\blacksquare$
Now it suffices to note that $(\ast)$ may be written in form $q(\alpha) = 0$, where
$$q(x) = x^2 - 2mx + m^2 - n$$
has rational coefficients and degree $2$.