Is it mathematically acceptable to use Prove if $n^2$ is even, then $n$ is even. to conclude since 2 is even then $\sqrt 2$ is even? Further more using that result to also conclude that $\sqrt [n]{2}$ is even for all n?
Similar argument for odd numbers should give $\sqrt[n]{k}$ is even or odd when k is even odd.
My question is does any of above has been considered under a more formal subject or it is a correct/nonsensical observation ?
Yes, essentially analogous arguments work to extend the notion of parity from the ring of integers to many rings of algebraic integers such as $\,\Bbb Z[\sqrt[n]{k}].\ $
The key ideas are: one can apply parity arguments in any ring that has $\ \mathbb Z/2\ $ as an image, e.g. the ring of all rationals with odd denominator, or the Gaussian integers $\rm\:\mathbb Z[{\it i}\,],\:$ where the image $\rm\ \mathbb Z[{\it i}\,]/(2,{\it i}-\!1) \cong \mathbb Z/2\ $ yields the natural parity definition that $\rm\ a\!+\!b\,{\it i}\ $ is even iff $\rm\ a\equiv b\ \ (mod\ 2),\ $ i.e. $ $ if $\rm\ a+b\,{\it i}\ $ maps to $\:0\:$ via the above isomorphism, which maps $\rm\ 2\to 0,\ i\to 1\:$.
Generally, it is easy to show that if $\rm\:2\nmid f(x)\in \Bbb Z[x]\setminus \Bbb Z\:$ then the number of ways to define parity in the ring $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(f(x))\ $ is given by the number of roots of $\rm\: f(x)\: $ modulo $2\:.\ $ For suppose there exists a homomorphism $\rm\ h\, :\, \mathbb Z[w]\to \mathbb Z/2.\:$ Then $\rm\:w\:$ must map to a root of $\rm\:f(x)\:$ in $\rm\ \mathbb Z/2.\ $ Thus if $\rm\ f(0)\equiv 0\ (mod\ 2)\ $ then $\rm\: \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ by $\rm\: x\mid f(x)\ (mod\ 2),\, $ and $\rm\, \!f(1)\equiv 0\ (mod\ 2) $ $\Rightarrow$ $\rm \mathbb Z[w]/(2,w\!-\!1) \cong \mathbb Z[x]/(2,x\!-\!1,f(x)) \cong \mathbb Z/2\, $ by $\rm\, x\!-\!1\,|\, f(x)\ (mod\ 2). $
Let's consider some simple examples. Since $\rm\ x^2\!+1\ $ has the unique root $\rm\ x\equiv 1\ (mod\ 2),\:$ the Gaussian integers $\rm\ \mathbb Z[{\it i}\,]\cong \mathbb Z[x]/(x^2\!+1)\ $ have a unique definition of parity, with $\:i\:$ being odd. Since $\rm\ x^2\!+x+1\ $ has no roots modulo $\: 2,\: $ there is no way to define parity for the Eisenstein integers $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+1).\, $ Indeed since $\rm\ w^3 = 1\ $ we infer that $\rm\: w \equiv 1\ (mod\ 2)\ $ contra $\rm\ w^2\!+w+1 = 0.\ $ On the other hand $\rm\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2\!+x+2)\ $ has two parity structures since both $\:0\:$ and $\rm\:1\:$ are roots of $\rm\ x^2\! + x + 2\ $ modul0 $\rm\:2,\:$ so we can define $\rm\:w\:$ to be either even or odd.
Note that that theorem begins with "Suppose $n$ is an integer, and that $n^2$ is even." So it does not hold when considering $\sqrt{2}$
Other answerers are answering several slightly different interpretations of your question; but taking literally what you asked,
Is it mathematically acceptable to use “Prove if $n^2$ is even, then $n$ is even.” to conclude since 2 is even then $\sqrt 2$ is even?
the answer is a quite interesting “no, but also yes”.
Formally, the answer is definitely no: as @john explained, you can’t use that theorem to conclude what you suggest, since the theorem and proof start by assuming that n is an integer, so $\sqrt{2}$ is not a valid value for $n$.
However, it is excellent and very acceptable mathematical practice to do what you did, and take that theorem as inspiration for considering a new generalisation of the concepts involved. And, as @KeyIdeas describes, this can lead to some very nice theories.
TL;DR: you can’t conclude this from that theorem/proof, but you can certainly be inspired along these lines by them.
The question you linked to starts (quite naturally) with "Suppose $n$ is an integer", which clearly excludes taking $n=\sqrt 2$. Therefore it cannot be used to conclude that $\sqrt2$ is even. Obviously $\sqrt2$ being even is absurd since even numbers in particular have to be integers. I would say your "observation" is sloppy (overlooking the stated conditions) and your following reasoning unfounded (or nonsensical if you prefer).
