= Solution
A nonempty compact <dynamical system> $(X,T)$ is a <minimal dynamical system> if it has no proper nonempty closed forward-invariant subset. Equivalently, every forward <orbit> is dense: the <orbit closure> of any point is a nonempty closed forward-invariant set, and conversely a closed invariant set contains the <orbit closure> of each of its points.
Consider the collection of nonempty closed forward-invariant subsets of $X$, ordered by reverse inclusion. Every chain has an upper bound, its intersection: the chain has the finite-intersection property, <compactness> makes the intersection nonempty, and continuity is not even needed for its forward invariance. The <Zorn lemma> supplies a minimal member $Y$, so $(Y,T|_Y)$ is a <minimal subsystem>. Also $T(Y)$ is compact, nonempty and forward invariant, so minimality gives $T(Y)=Y$. Here the printed word “nontrivial” must mean nonempty; a singleton with its identity map shows that a subsystem with more than one point cannot always be required.
If $y\in Y$, every tail of its forward <orbit> is dense in $Y$, since it is the orbit of $T^Ny\in Y$. Therefore every neighborhood of $y$ contains $T^ny$ for arbitrarily large $n$. Choosing these times successively for balls of radii $1/j$ gives $n_j\to\infty$ with $T^{n_j}y\to y$. This proves the <Birkhoff recurrence theorem>: \b[every <continuous map> of a nonempty compact metric space has a recurrent point]. In fact every point of the chosen <minimal subsystem> is recurrent.
Write the continuous circle extension as the <skew product>
$$
\widetilde X=X\times\mathbb R/\mathbb Z,\qquad\widetilde T(x,y)=(Tx,y+\rho(x)).
$$
A sufficient condition, and in this setting also a necessary one, is the <circle skew-product minimality criterion>: for every integer $m\ne0$ there is no continuous function $f:X\to\mathbb S^1$ such that
$$
\boxed{f(Tx)=e^{2\pi i m\rho(x)}f(x)\quad\hbox{for every }x.}
$$
The <cocycle> $\rho$ is assumed continuous; a merely measurable <cocycle> need not define this topological extension.
To prove sufficiency, choose a <minimal subsystem> $Z$ of the extension. Its projection is a nonempty closed forward-invariant subset of $X$, hence all of $X$. For $a\in\mathbb R/\mathbb Z$, vertical translation sends $Z$ to another <minimal subsystem>, since it commutes with $\widetilde T$. Two such translates either are disjoint or coincide: a nonempty intersection is closed and forward invariant, so minimality forces equality. Define their common stabilizer
$$
H=\{a:Z+a=Z\}.
$$
This is a closed subgroup of the <circle group>. Its closedness follows by taking limits of translations of points in the compact set $Z$, using both a translation and its inverse. Each fiber $Z_x$ is nonempty. If $y,z\in Z_x$, the sets $Z$ and $Z+(z-y)$ intersect at $(x,z)$, so $z-y\in H$. Conversely translating any fiber point by any member of $H$ stays in the fiber. Thus every fiber is exactly one coset of $H$.
If $Z$ is proper, $H$ is proper and therefore finite, say $H=\{0,1/m,\ldots,(m-1)/m\}$ for some positive integer $m$. The character $y\mapsto e^{2\pi imy}$ is constant on each fiber coset, giving a well-defined function $f(x)$ from $Z_x$. This function is continuous: its graph is the continuous image of $Z$ in $X\times\mathbb S^1$, and projection of that compact graph onto the Hausdorff space $X$ is a continuous bijection and hence a homeomorphism. The relation $\widetilde T(Z)=Z$ gives $f(Tx)=e^{2\pi im\rho(x)}f(x)$, contrary to the criterion. Hence $Z=\widetilde X$ and the extension is minimal.
Conversely, a solution of the displayed equation makes $F(x,y)=e^{2\pi imy}/f(x)$ a nonconstant continuous invariant function on the extension. Its nonempty level sets are proper closed invariant subsets, contradicting minimality. This also proves necessity of the <circle skew-product minimality criterion>.
Apply it with $\alpha=\sqrt2$, $X=\mathbb R/\mathbb Z$, $Tx=x+\alpha$, and $\rho(x)=2x+\alpha$. The base is a <minimal dynamical system>: the closure of the cyclic subgroup generated by an irrational angle is the whole <circle group>, and its positive forward orbit has the same closure. To see the subgroup assertion, a proper closed subgroup of the circle is finite; if it contained an irrational angle it would be infinite. The elementary classification follows because a closed subgroup with arbitrarily small nonzero elements is dense, while otherwise its least positive representative generates a finite subgroup.
There can be no forbidden $f$ for any $m\ne0$. Indeed, the <winding number of a continuous closed path> of $x\mapsto f(x+\alpha)$ equals that of $x\mapsto f(x)$, by translation and <homotopy invariance of winding number>. Consequently their quotient has <winding number> zero. But the proposed quotient $e^{2\pi im(2x+\alpha)}$ has <winding number> $2m\ne0$. Thus the <skew product> is minimal. Starting at $(0,0)$, induction gives
$$
\widetilde T^n(0,0)=(n\alpha,n^2\alpha)\pmod1,
$$
because $n^2\alpha+2n\alpha+\alpha=(n+1)^2\alpha$. Its <orbit> is dense in the two-dimensional <torus>, so its second coordinate is dense in the circle. Removing the initial point does not change density, since every orbit tail in a <minimal dynamical system> is dense. Therefore
$$
\boxed{\{n^2\sqrt2\pmod1:n\geq1\}\text{ is dense in }\mathbb R/\mathbb Z.}
$$
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