radicals - Prove $sqrt[m]{D}$ irrationality simply, without unique prime factorization

I apologize if this has been asked already. I am aware of proofs that $\sqrt[m]{D}$ is either integer or irrational for $m,D\in\mathbb{N}$, all of which that I recall and understand make use of the unique prime-factorization theorem (perfect example here; this one may not but I'm not familiar with the method). When I was young, I always disliked proofs that made use of unique prime factorization (for some reason) such as the standard proof of the irrationality of $\sqrt{2}$. However, there are other ways of proving $\sqrt{D}$ irrational, my personal favourite being the following proof (I have no idea where I got this from):

Let $D$ be a non-square integer. Suppose $\sqrt{D}$ is irrational and let $q$ be the smallest integer such that $q\sqrt{D}$ is integer, then for some $n\in\mathbb{N}$ we have:
$$n^2 $$\therefore\;\;n<\sqrt{D} $$\therefore\;\;0<\sqrt{D}-n<1$$
$$\therefore\;\;0 But $q\sqrt{D}-qn$ is an integer and so $(q\sqrt{D}-qn)\sqrt{D}=qD-qn\sqrt{D}$ is also an integer; thus $q\sqrt{D}-qn$ is an integer that contradicts minimality of $q$. Thus we have a contradiction and hence $\sqrt{D}$ is irrational.

I like this proof since it uses only basic algebra and does not invoke 'heavy' theorems (unless I'm missing something). I have wondered whether there is a proof of irrationality of $\sqrt[m]{D}$ that is spiritually related to this proof, i.e. one that just uses basic algebra without having to invoke unique prime factorization or other similar results of basic number theory, but I don't see a simple way of extending this proof.

Thus I have the following question: Does anyone know a simple proof of the irrationality of $\sqrt[m]{D}$ for $m\in\mathbb{N}$ and $D\ne k^m$ for any $k\in\mathbb{N}$ that does not rely on the unique prime-factorization theorem (or other similarly strong theorems), and preferably just uses basic algebra like the above proof for $m=2$?

Answer

I don't know if this is what is expected, anyway ...

Consider the equation $x^m-D=0$ (with $D\in\mathbb{N}$ not a perfect $m-$th power and $m\ge2$).

If $p/q$ is a rational root of this equation, with $(p,q)\in\mathbb{Z}\times\mathbb{N^\star}$ and $gcd(p,q)=1$,

then (by RRT) $q\mid 1$, so that $p/q$ is an integer, which is excluded.

Hence $\sqrt[m]{D}\not\in\mathbb{Q} \setminus \mathbb Z$.