Let $K/\mathbb{Q}$ be an abelian extension of degree $n$, and let $N_{K/\mathbb{Q}}(x)$ denote the norm of an element $x \in \mathcal{O}_K$. Then, with respect to a basis of $\mathcal{O}_K$ over $\mathbb{Z}$ as a $\mathbb{Z}$-lattice, one can interpret $N_{K/\mathbb{Q}}(x)$ as a homogenenous polynomial in $n$-variables, say $N(x) = N(x_1, \cdots, x_n)$. The form $N$ splits completely over $\overline{\mathbb{Q}}$ (in fact it already splits over $K$, since $K$ is abelian, thus Galois over $\mathbb{Q}$) and is irreducible over $\mathbb{Q}$. In this case it is possible to determine which integers $m$ may be written in the form $m = N(a_1, \cdots, a_n)$ for integers $a_1, \cdots, a_n$ in a similar fashion as in the case with binary quadratic forms (which are complete norm forms of degree 2). This is a consequence of class field theory. Of course, I cheated here, since like in the case of quadratic forms one has to take into account the situation when the class number exceeds one ($x^2 + y^2$ corresponds to $\mathbb{Q}(i)$, which has class number one). This can be generalized by replacing class field theory with say Chebotarev's density theorem.

In general, there does not seem to be an easy way to describe the numbers which can be written as $m = x^n + y^n$ for $n \geq 3$. It is conjectured (see for example: https://eudml.org/doc/153716) that for $n \geq 5$, for each integer $m$ such that $x^n + y^n = m$ has a solution in integers $x,y$, it has exactly two solutions, namely $(x,y)$ and $(y,x)$. In the language of my paper with C.L. Stewart (https://arxiv.org/abs/1605.03427), this asserts that each $m$ that can be so represented is *essentially* represented.

In that joint paper with Stewart I mentioned above, we showed that for any binary form $F$ with integer coefficients, non-zero discriminant, and degree $d \geq 3$ there exists a positive number $C_F$ such that the number of integers $m$ with $|m| \leq X$ and for which the Thue equation $F(x,y) = m$ has a solution is asymptotic to $C_F X^{2/d}$. For $d \geq 3$ this set is too thin to have a description like in the quadratic case (any such set has density like $X (\log X)^{-a}$ for some explicit rational number $a > 0$). There are a lot of subtleties with this set. For example, even for the binary cubic form $xy(x+y)$, the existence of an infinite sequence $\{m_s\}$ of cube-free integers such that the number of solutions to the equation $xy(x+y) = m_s$ tends to infinity as $s \rightarrow \infty$ would imply that the Mordell-Weil rank of elliptic curves is unbounded, which seems to be an ever more fantastical situation.

I should add that the projective curve

$$\displaystyle x^n + y^n = mz^n$$

has complex multiplication, and hence one can in principle determine the set of rational points on the curve via Minhyong Kim's non-abelian Chabauty method, as shown by Coates and Kim in this paper: https://arxiv.org/abs/0810.3354