There are many functions $f(x,y)$ such that the equation:
$$x^3 + y^3 + (f(x,y))^3 = (f(x,y)+1)^3\tag1$$
has isolated solutions in integers, not forming cycles except in the trivial sense that reversing $x$ and $y$ forms a 2-cycle. For example, if $f(x,y)=x+y-2$, then the only solutions of (1) are $(3,4), (4,3), (0,1), (1,0)$. Note that any solution must have $x$ and $y$ of opposite odd-even parity.
Underlying the cycles described in the question is the case in which, for some integer $k$:
$$f(x,y) = 3k^2(x+y-1)+y-1$$
In that case, (1) is equivalent to:
$$x^3 + y^3 – 27k^4x^2 – (27k^4+18k^2+3)y^2 – (54k^4+18k^2)xy + (54k^4+9k^2)x + (54k^4+27k^2+3)y – (27k^4+9k^2+1)=0\tag2$$
I. A first simplication is that this factorises as:
$$(x+y-1)\big[x^2 - xy + y^2 - (27k^4-1)x - (27k^4+18k^2+2)y +(27k^4+9k^2+1)\big] = 0\tag3$$
Equating the first factor to zero leads only to trivial solutions, but equating the second to zero yields the equation of an ellipse. Using notation from Wikipedia, this can be written as:
$$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\tag4$$
where:
$\qquad A = 1,\; B = -1,\; C = 1$
$\qquad D = -(27k^4-1)$
$\qquad E = -(27k^4+18k^2+2)$
$\qquad F = 27k^4+9k^2+1$
and $D+E+2F = 1.\,$ Since $x,y$ must be of opposite parity, then $Ax^2 + Bxy + Cy^2$ is odd, also $F$ is odd regardless of the parity of $k$. So a solution of (4) must have $Dx+Ey$ even, which requires either odd $(k,x)$ and even $y$, OR even $(k,x)$ and odd $y$.
II. A second simplification is that from any one integer solution of (4) we can obtain another. To see this, suppose we have a solution in integers $k$ (odd), $x_0$ (odd) and $y_0$ (even). Substituting $y_0$ only into (4) we have a quadratic equation in $x$:
$$Ax^2 + (By_0+D)x + (Cy_0^2+Ey_0+F) = 0\tag5$$
This will normally have two roots, one of which must be the known $x_0$. To obtain the other – call it $x_1$ – we can use the fact that the sum of the roots of an equation $ax^2+bx+c=0$ is $-b/a$, so:
$$x_1 = -\dfrac{By_0+D}{A}-x_0\tag6$$
Since $B,y_0,D$ are integers while the denominator $A=1$, $x_1$ must also be an integer, so (barring degenerate cases where $x_1=x_0$), then $(x_1,y_0)$ is a new solution in integers of (4). This reasoning remains valid for any integer value of $k$.
As an illustration, if $k=1$ and $(x_0,y_0)=(-9,16)$, equation (5) becomes:
$$x^2 + \big(-1(16) –(27-1) \big)x + \big(16^2 – (27+18+2)16 + (27 + 9 +1 \big) = 0\tag7$$
which simplifies to:
$$x^2 -42x - 459 = 0\tag8$$
yielding $(x_1,y_1)=(51,16)$ as a new solution of (4). The two solutions are at the intersections of the ellipse and the horizontal line $y=16$ as shown in Chart 1.
This procedure can be repeated by substituting $x_1$ into (4) and solving in the same way the resulting quadratic equation in $y$ to obtain a further integer solution of (4) which will lie on the intersection of the ellipse with a vertical line through $(x_1,y_0)$. And there is no need to stop there: we can proceed indefinitely, at each iteration substituting into (4) the new value of $x$ or $y$ obtained at the previous iteration. Starting from $(-9,16)$, this yields the sequence of solutions $(-9,16),(16,51),(51,82),(82,57),(57,22),(22,-9),\dots$.
A simple argument shows that this procedure must eventually result in a complete cycle. There are only a finite number of lattice points (points with integer coordinates) in the vicinity of an ellipse defined by equation (4), for example the whole of Chart 1 contains 121 such points. The infinite series of integer points generated as above, all lying on the ellipse (and, a fortiori, within that vicinity) must therefore eventually revisit at least one point.
III. A third consideration is that the configuration of the ellipses defined by equation (4) is conducive to cycles of 6 points, rather than some other number of points. Intuitively, the number of points that can form a cycle linked by alternate horizontal and vertical lines within an ellipse is related to a) the direction of the axes, and b) its eccentricity, one indicator of which is the ratio of the major to the minor axis.
If the axes are horizontal and vertical, a 4-cycle is always possible as in Chart 2 below. Note that it is not claimed that any ellipse must pass through at least one integer point, nor that if it does then another such point must lie on the intersection of the ellipse with a horizontal or vertical line throught the first point. The purpose of Charts 2 and 3 is to show that, if those conditions are met, the number of points in the resulting cycles will depend on the configuration of the ellipse.
If the axes are at $45^{\circ}$ to the horizontal, then a high degree of eccentricity will require a cycle with a large number of points, as in Chart 3 below in which the ratio of the axes is about $6$:
The ellipses defined by equation (4) are moderately eccentric as shown in Chart 1. This is true whatever the value of $k$ since the ratio of major to minor axis is independent of $k$, being given by (see Wikipedia, formula for a,b under General Ellipse):
$$\sqrt{\Bigg(\dfrac{A+C+\sqrt{(A-C)^2+B^2}}{A+C-\sqrt{(A-C)^2+B^2}}\Bigg)}=\sqrt{\Bigg(\dfrac{1+1+\sqrt{(1-1)^2+(-1)^2}}{1+1-\sqrt{(1-1)^2+(-1)^2}}\Bigg)}=\sqrt{3}$$
I am not aware of a precise formula relating the cycle size to the degree of eccentricity of an ellipse with axes at $45^{\circ}$ to the horizontal, but having seen the example of the ellipse in Chart 3 it seems unsurprising that ellipses with an axis ratio of $\sqrt{3}$, ie moderately eccentric, should generate 6-cycles, two of which from the question are illustrated below:
Each of the two cycles is the mirror image of the other, the axis of symmetry being the line $y=x+7$ which coincides with the major axis of the ellipse.
