From the second edition of Unsolved Problems in Number Theory by Richard K. Guy, section D1, Euler conjectured that your equation should be impossible for $n \geq 4.$ For $n=4,$ the first counterexample was by Noam Elkies, a parametric solution was given by Dem'janenko, and your solution above was found by Roger Frye. No explicit reference for Frye...There is a third edition, there it starts on page 209. I don't think anyone has any solutions for $n>4.$
To be specific, Euler thought that you needed four fourth powers to add up to another fourth power, five fifth powers to add up to another fifth power, and so on.
OK, we have
$$ 27^5 + 84^5 + 110^5 + 133^5 = 144^5, $$
by Lander and Parkin (1966). I don't think three fifth powers is going to work. Might try six or five or four sixth powers, or seven or six or five seventh powers if you want to stick with prime exponents. Note that, by Fermat's little theorem, for sixth powers , all but one of the summands will need to be divisible by 7. That is, if some $a \neq 0 \pmod 7$ then $a^6 \equiv 1 \pmod 7.$ You have fewer than seven summands, so the left hand side will not be divisible by 7, so the right hand side will be exactly $1 \pmod 7.$ Something of a timesaver. To compare, look at the solution in the original question for $n=4,$ two out of three numbers on the left hand side are divisible by 5.
Alright, as of the second edition (1994) there were no known solutions for $n$ $n$th powers adding up to another $n$th power for $n \geq 6.$ So, even what Euler had intended as the problem had no known solutions. Of course, computers are much faster now...
I will try to give you a VERY quick overview on the strategy of the proof of FLT. Of course I cannot avoid to use very technical tools, such Galois representation. If you don't know about these stuff, I hope you can at least follow the "shape" of the argument.
The starting observation is the following: for $n\in\mathbb N$, let FLT($n$) be the statement "there are no triples $(a,b,c)$ of integers with $abc\neq 0$ such that $a^n+b^n=c^n$". Then it is elementary to see that FLT($d$) implies FLT($n$) whenever $d\mid n$. Therefore it is sufficient to prove FLT($p$) for every odd prime $p$ and for $n=4$ in order to prove FLT($n$) for all positive integers $n\geq 3$. The cases $p=3$ and $n=4$ were already known to Euler, I believe, and can be proven by elementary methods, so we can assume that $p\geq 5$.
Now the non-elementary math comes in. Suppose that there is a triple of integers $(a,b,c)$ which contradicts FLT($p$) for some prime $p$. One can construct the following elliptic curve over $\mathbb Q$:
$$E_{a,b,c}\colon y^2=x(x-a^p)(x-b^p)$$
It is possible to show (assuming wlog that $(a,b,c)$ is a coprime triple, $a\equiv -1\bmod 4$ and $2\mid b$) that this is a semistable elliptic curve whose conductor is $\prod_{l\mid abc}l$.
Moreover, Serre and Frey proved the following theorem: let
$$\overline{\rho}_{a,b,c}\colon \text{Gal}(\overline{\mathbb Q}/\mathbb Q)\to GL_2(\mathbb F_p)$$
be the residual Galois representation at $p$ attached to $E_{a,b,c}$. Then:
- $\overline{\rho}_{a,b,c}$ is absolutely irreducible;
- $\overline{\rho}_{a,b,c}$ is odd;
- $\overline{\rho}_{a,b,c}$ is unramified outside $2p$ and flat at $p$
The key idea is then the following: suppose that we can show that for every (semistable) elliptic curve $E$ over $\mathbb Q$ the representation $\overline{\rho}_E$ is the residual representation of the $p$-adic Galois representation attached to a weight $2$ newform $f_E$. Then we can apply a theorem of Ribet, using the properties of $\overline{\rho}_{a,b,c}$, to show that for $\overline{\rho}_{a,b,c}$ we can choose such a newform in $S_2(\Gamma_0(2))$. But then we have a big problem: this space is $0$-dimensional! Therefore we are led to a contradiction, and there cannot be any triple $(a,b,c)$ contradicting FLT($p$).
This whole argument was already known long before Wiles' proof. But the thing that was missing was the proof of the (nowaday) so-called modularity theorem: every elliptic curve over $\mathbb Q$ is modular, i.e. its $p$-adic Galois representation is the $p$-adic Galois representation attached to a weight $2$ newform.
Even though this was the only missing part, it is by far the most difficult one! Wiles was able to prove it for semistable curves, which was enough for proving FLT, but some years later the result has been improved to all elliptic curves over $\mathbb Q$ by Breuil, Conrad, Diamon and Taylor.
Anyway, if you are interested in more details, you can read the (amazing) first chapter of "Modular forms and Fermat's last theorem", by Cornell, Silverman and Stevens eds. Of course it is a math textbook, so you need some background in these type of topics in order to understand it.
Best Answer
On the other hand, you can do the same by noting that $\sqrt[n]{2}$ is a root of
$$x^n - 2 = 0$$
and rational root theorem says that the only rational roots, if any, must be one of
$$\{-1, 1, -2, 2\}$$
none of which are proper candidates for $n > 1$, so $\sqrt[n]{2}$ is not rational.
Edit : Further, you can just use Gauss's lemma by noting that
$$x^n - 2$$
is irreducible over $\mathbb{Z}[x]$, as the sign changes occur in the interval $(2, 1)$ and the negative counterpart assuming for all $n > 1$, none of which contains integers, and thus by Gauss's lemma is also irreducible over $\mathbb{Q}[x]$.