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.
We get $z(y^2-z)=x^3$. Thus $z=ab^2c^3$, $y^2-z=ba^2d^3$ for certain integers $a, b, c, d$ (that is easy to see considering the prime factorization). So $ab(bc^3+ad^3)=y^2$. Denote $a=TA$, $b=TB$ (each pair $(a, b) $ corresponds to at least one triple $(A, B, T)$, but possibly to several triples). You get $T^3AB(Bc^3+Ad^3)=y^2$.Thus $T$ divides $y$, say $y=TY$. We get $Y^2=TAB(Bc^3+Ad^3)$.
So, all solutions are obtained as follows: start with arbitrary $A, B, c, d$ and choose any $Y$ which square is divisible by $AB(Bc^3+Ad^3)$, the ratio is denoted by $T$ (if both $Y$ and $AB(Bc^3+Ad^3)$ are equal to 0, take arbitrary $T$).
Best Answer
It does have infinitely many positive solutions. Here is just one such series.
Consider the following recurrence sequence:
$$u_0=1,\ u_1=2,\ u_{n+1} = 23 u_n - u_{n-1} - 4\qquad (n\geq 1).$$
Let $t,k$ be any two consecutive terms of this sequence, then setting $l:=k^2+t$ produces the following equality: $$(3+3k+l)(t+1) = (k+26)(kl-k^3-1),$$ which gives solution $m:=\frac{(k+26)(3+3k+l)}{t+1}$ (which is an integer) to the original equation.
In fact, integrality of $m$ follows from the identity: $$(u_{n+2}+1)(u_n+1) = (u_{n+1}+26)(u_{n+1}+1),$$ which can be verified from the recurrence for $u_n$.
In summary, the values $(k,l,m)$ in this solution series are given by $$\begin{cases} k = u_{n+1}, \\ l = u_{n+1}^2 + u_n, \\ m = (u_{n+2}+2)(u_{n+1}+2) + 24. \end{cases}\qquad (n\in\mathbb{Z}_+) $$
ADDED. I've added $u_n$ to the OEIS as sequence A350917. Together with 9 other similar recurrences it gives all solutions $k$ to $(tk-1)\mid (k+1)^4$, which are now listed in sequence A350916.