Revised
Interesting question.
Here's a thought:
You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms
$x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the
multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times
the order 2 group generated by $-1$.
If you take a submonoid that uses only
one prime, it is quasi-isometric to a quotient of the hyperbolic plane by an action of $\mathbb Z$, which is multiplication by $p$ in the upper half-space model. To see this, place a dots labeled by integer $n$ at position $(n*p^k, p^k)$ in the upper half plane, for every pair of integers $(n,k)$, and connect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The quotient of upper half plane by the hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the Cayley graph for this monoid. This is also quasi-isometric to the 1-point union of two copies of the hyperbolic plane, one for negative integers, one for positive integes. It's a fun exercise,
using say $p = 2$. Start from 0, and recursively build the graph by connecting $n$ to $n+1$ by one color arrow, and $n$ to $2*n$ by another color arrow. If you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the corresponding graph for a different prime.) The negative integers look just the same, but with the successor arrow reversed.
If you use several primes, the picture gets more complicated. In any case, one can take rescaled limits of these graphs, based at sequence of points, and get asymptotic cones for the monoid. The graph is not homogeneous, so there is not just one limit.
Another point of view is to take limits of $\mathbb Z$ without rescaling, but
with a $k$-tuple of constants $(n_1, \dots , n_k)$. The set of possible identities among polynomials in $k$ variables is compact, so there is a compact space of limit rings for $\mathbb Z$ with $k$ constants. Perhaps this is begging the question: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions.
Rescaling may eliminate some of this complexity.
A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to
$\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on some asymptotic cone for the affine monoid for $\mathbb Z$.
With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.
A note on the observation "$n$ good implies $2n + 2$ good":
First remark is that $n$ is good iff there are positive rational numbers $a_1, \dotsc, a_m$ such that $n = (a_1 + \dotsc + a_m)(1/a_1 + \dotsc + 1 / a_m)$. This is because one can multiply all $a_i$ by the lcm of their denominators.
Second remark is that $n$ is good iff there are positive rational numbers such that $n = 1/a_1 + \dotsc + 1/a_m$ and $a_1 + \dotsc + a_m = 1$. This is because one can multiply all $a_i$ by the inverse of their sum.
Finally, if $n = 1/a_1 + \dotsc + 1/a_m$ with $a_1 + \dotsc + a_m = 1$, then taking $a_i' = a_i / 2$ for $1\leq i \leq m$ and $a_{m + 1}' = 1/2$ gives us $2n + 2$.
Looking at the proof by Graham cited above, I see that this observation is indeed halfway to the actual proof. Namely we need a second result that $n$ good implies $2n + 179$ good, to take care of the odd integers. Then the result follows from the fact that all integers from $78$ to $333$ are good.
Since we don't require the $a_i$'s to be different, we may take $a_i' = a_i / 2$ for $1 \leq i \leq m$, $a_{m + 1}' = 1/3$, $a_{m + 2}' = 1/6$, and get $2n + 9$ good.
This also suggests that we might improve significantly the bound of Graham.
Following the comment of Max Alekseyev, this OEIS sequence says that all positive integers except $2, 3, 5, 6, 7, 8, 12, 13, 14, 15, 19, 21, 23$ are Egyptian, and hence good.
Since our definition of "good" is weaker than "Egyptian", it doesn't a priori exclude the possibility that some of the above listed numbers are still good.
Therefore it only remains to see whether the above numbers are good.
The task is separated into $m = 2, 3, 4$.
For $m = 2$, it is clear that only for $n = 4$, the equation $(x + y)(1/x + 1/y) = n$ has rational solution $(x, y)$.
For $m = 3$, we are led to the equation $(x + y + z)(xy + yz + zx) = nxyz$. Assuming $n > 9$ and choosing the point $[x, y, z] = [-1, 1, 0]$ as the point at infinity, we get an elliptic curve.
A Weierstrass equation is given by $Y^2 = X^3 + (n^2 - 6n - 3)X^2 + 16nX$, where $X = \frac{4n(x + y)}{x + y - (n - 1)z}$, $Y = \frac{4n(n - 1)(x - y)}{x + y - (n - 1)z}$. To get back $[x, y, z]$ from $X, Y$, we have the formula $[x, y, z] = [(n - 1)X + Y, (n - 1)X - Y, 2X - 8n]$.
This curve has six torsion points, which correspond to useless solutions to our equation. Moreover, the Mordell-Weil group has rank $0$ for all the above $n$, except $n = 14, 15$. This proves that $n = 12, 13$ are not good.
For $n = 14$, we have a solution $(3, 10, 15)$.
For $n = 15$, we have a solution $(1, 3, 6)$.
Thanks again to Max Alekseyev, $n = 19, 21, 23$ are all good, with solutions $(5,8,12,15), (8,14,15,35), (76,220,285,385)$, respectively.
Conclusion: A positive integer is good if and only if it is not one of $2, 3, 5, 6, 7, 8, 12, 13$.
Best Answer
Seeing that the link is to a 1996 announcement that I posted to sci.math.research, I suppose I should explain. Yes, the formulation in that announcement is not clearly stated $-$ one doesn't polish USENET posts like a published paper, and can't even edit after the fact (as is possible on mathoverflow) to correct blatant typos like the stray "+1" in the formula "(x,y,z)=(1+6t^3+1,1-6t^3,-6t^2)".
As it happens here there is a published paper that appeared only a few years later:
but the relevant section (3.2) doesn't address parametrizations of $x^3+y^3+z^3=2$. The answer to the present question is that D.Burde is basically right: the correct statement was, and still is, that all known solutions of $x^3 + y^3 + z^3 = 2$ in ${\bf Q}[t]$ come from the identity $$ (1+6t^3)^3 + (1-6t^3)^3 + (-6t^2)^3 = 2 $$ by permuting $x,y,z$ and substituting some polynomial for $t$. The substitution need not be linear, but nonlinear substitutions like $(x,y,z) = (1+6t^6, 1-6t^6, -6t^4)$ give no new $(x,y,z)$ solutions either. I don't think any method is known that would prove that there are no other nonconstant solutions, or that there's no nonconstant solution in ${\bf Q}[t]$ to $x^3+y^3+z^3=d$ unless $d$ is a cube or twice a cube. All that can be said is that if there were such a solution that had small enough degree and coefficients then it would have turned up in searches for integral solutions such as the searches described in that ANTS-4 paper and also on this page.