Yes, your question is closely related to rational varieties.
As the article you link to and Cass' comment explain, a rational variety $X$ of dimension $n$ is one that has a dense open subset that is isomorphic to an open subset of affine space $\mathbf A^n$. This isomorphism allows us to identify the coordinate functions on $X$ with rational functions on $\mathbf A^n$, just as in your example. So this gives a rational parametrisation of the points on $X$.
Conversely, if you have a rational parametrisation of the points of $X$, plugging in values of the parameters gives a dominant rational map $f:\mathbf A^m \dashrightarrow X$. (We don't demand that parametrisations be defined for all possible values of the parameters, just a dense open set of the possible values. Note also that our parametrising affine space is allowed to have bigger dimension than $X$.)
A variety $X$ for which one can find a map $f$ as described above is called unirational. It's an easy exercise to see that if there is a dominant rational map as above, then one can find a dominant map $\varphi :\mathbf A^n \dashrightarrow X$ from an affine space of the same dimension as $X$. Such a map must be generically finite, so it gives a finite-to-one parametrisation of the points of $X$.
However: unirational does not imply rational! It was known by around 1900 that a curve or surface that has a finite-to-one parametrisation must actually have a generically one-to-one (in other words birational) parametrisation. The question of whether this remains true in higher dimensions remained open until the early 1970s, when, remarkably, three completely independent proofs (by Artin–Mumford, Clemens–Griffiths, and Iskovskikh–Manin) showed that a smooth cubic hypersurface in $\mathbf P^4$ is unirational but not rational.
In general, deciding whether varieties are rational or not remains one of the most fundamental and difficult issues in algebraic geometry.
On the other hand: although there are some cases in which deciding rationality is extremely difficult, the comment of Cass that varieties are "rarely" rational is quite accurate, because there are easliy-verified necessary conditions for rationality that rule out "most" candidates.
For example, let $X$ be a smooth hypersurface in $\mathbf P^n$. Then the adjunction formula shows that if $X$ has degree at least $n+1$, then the canonical bundle $\omega_X = \wedge^{n-1} \Omega_X$ has nonzero global sections. This is impossible for a rational variety, and so we conclude that no smooth hypersurface of degree at least $n+1$ is rational.
Let $k$ be any field and let $B:=k[x_1,..,x_n]$ be the polynomial ring in $n$ variables. Let $I \subseteq B$ be an ideal with $A:=B/I$ and let $X:=Spec(A)$ and let $S:=Spec(k)$.
Let
R0. $X(k)$ be the set of $n$-tuples $(a_1,..,a_n)\in k^n$ with $f(a_1,..,a_n)=0$ for all polynomials $f \in I$.
R1. Let $Hom_{k-alg}(A,k)$ be the set of all maps of $k$-algebras $\phi: A \rightarrow k$.
Lemma. There is a one-to-one correspondence of sets $X(k) \cong Hom_{k-alg}(A,k)$.
If $a:=(a_1,..,a_n)\in X(k)$ we define a map $\phi_a: A\rightarrow k$ by letting $\phi_a(x_i):=a_i$. Conversely if $\phi: A\rightarrow k$ is a map of $k$-algebras it follows $\phi(x_i)=a_i \in k$ for all $i=1,..,n$. It follows $a:=(a_1,..,a_n)\in X(k)$. This is a one-to-one correspondence of sets.
People write $X(k)$ to denote the "set of $k$-rational points of $X$". A $k$-rational point is by definition a map of schemes
R2. $\phi^*: Spec(k) \rightarrow X$
which again corresponds 1-1 to a map of $k$-algebras $\phi:A \rightarrow k$. You do not need $k$ to be a field, you can let $k$ be any commutative unital ring. By the construction of the affine scheme $Spec(A)$ it follows
R3. $Hom_{Sch(S)}(S,X)\cong Hom_{k-alg}(A,k)$
Example 1. Let $I:=\{f_1,..,f_l\}$ be generated by a finite set of polynomials.
It follows a map of schemes $\phi^*: S\rightarrow X$ corresponds in a 1-1 way to a solution $(a_1,..,a_n)\in k^n$ of the system of polynomial equations
S1. $f_1(x_1,..,x_n)=0,..,f_l(x_1,..,x_n)=0$.
Hence the "functor of points" language converts the problem of studying sets of solutions to systems of polynomial equations into the problem of studying sets of maps $\phi^*: S \rightarrow X$.
Question: "Sorry I do not know this notation, what do you mean by Spec(A)?"
Answer: In chapter II in Hartshornes (or any book on algebraic geometry) book the ringed topological space $Spec(A)$ is introduced. Given any commutative unital ring $A$ you construct a topological space $X:=Spec(A)$ (it is the set of prime ideals in $A$ with the Zariski topology) and a structure sheaf $\mathcal{O}_X$. Given any map of unital commutative rings $\phi: A\rightarrow B$ (let $Y:=Spec(B)$) you get a map of locally ringed spaces
$(\phi^*, \phi^{\#}): (Y, \mathcal{O}_Y) \rightarrow (X, \mathcal{O}_X)$
with the property that the map of ringed topological spaces $(\phi^*, \phi^{\#})$ is uniquely determined by $\phi$. There is a 1-1 correspondence
R4. $Hom_{Sch}(Y,X) \cong Hom_{rings}(A,B)$.
This is Proposition II.2.3 in Hartshornes classical book "Algebraic geometry".
Hence you get a "geometric structure" $(X, \mathcal{O}_X)$ from any commutative unital ring $A$. And to study a map of schemes $\phi^*: S \rightarrow X$ is by the above argument equivalent to studying a solution to the system S1 from Example 1.
Why study solutions to systems of polynomial equations over rings that are not fields?
Example. In diophantine geometry one studies integral solutions to systems of polynomial equations. Given a set of polynomials $I:=\{f_1,..,f_l\}\in B:=\mathbb{Z}[x_1,..,x_n]$ we may construct the quotient ring $A:=B/I$. Let $S:=Spec(\mathbb{Z})$ and $X:=Spec(A)$. It follows there is a 1-1 correspondence of sets
R5 $Hom_{Sch}(S,X)\cong Hom_{rings}(A, \mathbb{Z})\cong X(\mathbb{Z})$.
For any scheme $X\in Sch(S)$ there is the "functor of points"
R6. $h_X: Sch(S) \rightarrow Sets$
where $Sch(S)$ is the category of schemes over $S$ and maps of schemes over $S$
and $Sets$ is the category of sets and maps of sets. The functor $h_X$ is defined by
R7. $h_X(T):=Hom_{Sch(S)}(T,X)$.
This gives an "embedding"
R8. $h: Sch(S) \rightarrow Funct(Sch(S), Sets)$.
Here $Funct(Sch(S), Sets)$ is the "category of functors" from $Sch(S)$ to $Sets$ with natural transformations as morphisms.
It is an "embedding" in the sense that any natural transformation $\eta:h_X \rightarrow h_Y$ of functors is induced by a unique morphism of schemes $f:X\rightarrow Y$, and two different morphisms $f,g:X\rightarrow Y$ give different natural transformations (this is the Yoneda lemma). With the functor $h$ we may view the category $Sch(S)$ as a "sub category" of $Funct(Sch(S), Sets)$. This viewpoint is what people use to define algebraic spaces, algebraic stacks and other "moduli spaces/parameter spaces" in algebraic geometry. As an example:
Let $E$ be a rank $d+1$ locally trivial $\mathcal{O}_S$-module and define the following functor:
R9. $F_{E}: Sch(S) \rightarrow Sets$
by $F_{E}(X,f):= \{\phi: f^*E^* \rightarrow L \rightarrow 0$ such that the sequence is exact $\}/\equiv$
where $L\in Pic(S)$ and two quotients $(L, \phi), (L', \phi')$ are equivalent iff there is an isomorphism $\psi: L \rightarrow L'$ such that the obvious diagram commutes. The functor $F_{E}$ is in the "category" $Funct(Sch(S), Sets)$, and it is in the "image" of the functor $h$: There is a scheme $\mathbb{P}(E^*)$ (the projective space bundle of $E$) and an isomorphism
of functors $h_{\mathbb{P}(E^*)}\cong F_{E}$. We say that the functor $F_E$ is "representable" and that "it is represented by $\mathbb{P}(E^*)"$. You may find this in Hartshorne, Chapter II.7, Proposition 7.12. Many functors
appearing in the study parameter spaces are not representable in this sense, and this leads to the theory of algebraic spaces and algebraic stacks.
"Many" "moduli problems" may be formulated as follows: Given a functor
$F: Sch(S) \rightarrow Sets.$
Question: Is there a scheme $f: X\rightarrow S$ with the property that $F(-) \cong h_X$ is an isomorphism of functors? Projective space bundles, grassmannian bundles, quot schemes etc are constructed using this language.
If $E$ is a locally trivial $\mathcal{O}_S$-module of rank $d+1$ you may construct the projective space bundle $\pi: \mathbb{P}(E^*)\rightarrow S$. There is the "tautological sequence"
T1. $\pi^*E^* \rightarrow \mathcal{O}(1) \rightarrow 0$.
If $S=k$ is a field it follows $E=V$ is a vector space of dimension $d+1$. If you dualize T1 you get an injection
T2. $\mathcal{O}(-1) \subseteq \pi^*E$
Given a $k$-rational point $x:Spec(k) \rightarrow \mathbb{P}(V^*)$
you get a line
T3. $\mathcal{O}(-1)(x) \subseteq V$ and this is a one to one correspondence.
This reflects that projective space is a parameter space: It parametrizes lines in $V$.
https://en.wikipedia.org/wiki/Moduli_space
There is the "Yoneda lemma":
https://en.wikipedia.org/wiki/Yoneda_lemma
Best Answer
The notion of a $K$-rational point depends very much on the field $K$. For example, consider the algebraic variety given as the zero set of the polynomial $x^2+y^2+1\in \mathbb{C}[x,y]$. Note that this variety has no $\mathbb{R}$-rational points.
Exercise 1: Prove that if we consider the algebraic variety given as the zero set of the polynomial $x^2+y^2-1\in \mathbb{R}[x,y]$, then this variety has infinitely many $\mathbb{Q}$-rational points.
In light of Exercise 1, note that the set of $\mathbb{Q}$-rational points of this variety is a proper subset of the set of $\mathbb{R}$-rational points of this variety.
In general, if $X$ is a scheme over a field $K$, then we can speak of the set of all $K$-rational points of $X$. Note that the $K$-rational points of $X$ are precisely those points of $X$ at which the residue field of the local ring at that point is isomorphic to $K$.
Definition: Let $\pi:X\to S$ be an $S$-scheme. A $\textit{section}$ of $X$ is a morphism of $S$-schemes $\sigma:S\to X$. This amounts to saying that $\pi\circ\sigma=\text{Id}_S$. The set of sections of $X$ is denoted by $X(S)$ (and also by $X(A)$ if $S=\text{Spec}(A))$).
Exercise 2: Prove that if $X$ is a scheme over a field $K$, then we can identify $X(K)$ with the set of all $K$-rational points of the scheme $X$.
Let $K$ be a field and let $X=\text{Spec }K[T_1,\dots,T_n]/I$ be an affine scheme over $K$. Let $Z$ be the zero set of some polynomial $P\in I$. It is natural to ask whether our notion of a "$K$-rational point of $X$" in the scheme-theoretic sense agrees with our naive notion of a $K$-rational point of the algebraic set defined by $P$.
Exercise 3: Prove that we have a canonical bijection $Z\to X(K)$ where $X(K)$ denotes the set of all $K$-rational points of the scheme $X$.