I'm not an expert here, but reading J.S.Milne's notes on elliptic curves (available from his website), I noticed the following statement.
ASIDE 2.4 There is a heuristic explanation for Mordell’s conjecture. Let $C$ be a curve of genus $g\ge1$ over $\mathbb{Q}$, and assume that $C(\mathbb{Q})\not=\emptyset$. It is possible to embed $C$ into another projective variety $J$ of dimension $g$ (its jacobian variety). The jacobian variety $J$ is an abelian variety, i.e., it has a group structure, and a generalization of Mordell’s theorem (due to Weil) says that $J(\mathbb{Q})$ is finitely generated. Hence, inside the $g$-dimensional set $J(\mathbb{C})$ we have the countable set $J(\mathbb{Q})$ and the (apparently unrelated) one-dimensional set $C(\mathbb{C})$. If $g > 1$, it would be an extraordinary accident if the second set contained more than a finite number of elements from the first set.
Now, I'm not completely convinced about quite how unrelated the one-dimensional set $C(\mathbb{C})$ and the countable set $J(\mathbb{C})$ are. More generally, Lang's conjecture (now a theorem) says that if $X$ is a subvariety of an abelian variety $A$ and $\Gamma$ is a finite rank subgroup of $A(\mathbb{C})$, then the Zariski closure of $X(\mathbb{C})\cap\Gamma$ is a union of finitely many translates of subvarieties of $A$. In particular, if $X$ does not contain a translate of a nontrivial abelian subvariety of $A$, then $X(\mathbb{C})\cap\Gamma$ is finite (I'm working from the statement of Lang's conjecture given in Theorem F.1.1.1 of Diophantine Geometry: An Introduction by Hindry and Silverman). So, it does seem that subvarieties and finitely generated subgroups of abelian varieties have an intersection which is small.
I'm also going to add an elementary heuristic argument to try and answer James Borger's question above, "Why should a generic polynomial $f(x,y)\in\mathbb{Q}[X,Y]$ of degree $d\ge4$ have only finitely many solutions in $\mathbb{Q}$?". This is something I have thought about before, and can at least come up with a rough heuristic argument why this should be the case. Setting $g(x,y,z)=f(x/z,y/z)z^d$, which is a homogeneous polynomial in $\mathbb{Q}[x,y,z]$, the question is equivalent to asking why $g$ should have only finitely many coprime solutions in $\mathbb{Z}^3$.
The probability of a randomly chosen triple $(x,y,z)\in\mathbb{Z}^3$ being coprime is $\zeta(3)^{-1}$. So, in the ball $B_R$ of radius $R$ centered about the origin in $\mathbb{R}^3$, we expect about $\frac43\pi R^3\zeta(3)^{-1}=O(R^3)$ such triples. We want to know how many satisfy $g(x,y,z)=0$. Choosing any $a\gg1$, we can instead ask how many coprime triples $(x,y,z)\in\mathbb{Z}^3$ satisfy $\vert g(x,y,z)\vert\le a$. Letting $V$ be the surface $\{{\bf x}\in\mathbb{R}^3\colon g({\bf x})=0\}$, we can approximate $\vert g({\bf x})\vert$ to leading order as$$\vert g({\bf x})\vert\approx \Vert\nabla g({\bf x})\Vert d({\bf x},V),$$where $d({\bf x},V)$ is the distance of a point ${\bf x}\in\mathbb{R}^3$ from $V$. This gives the following estimate for the volume of the set ${\bf x}\in B_R$ with $\vert g({\bf x})\vert\le a$ in terms of an integral over the surface $V$,$$2a\int_{V\cap B_R}\frac{1}{\Vert \nabla g({\bf x})\Vert}\,d{\bf x}.$$As $\nabla g$ is homogeneous of degree $d-1$, this can be rewritten as a line integral over the intersection of $V$ with the unit sphere $S^2$,$$2a\int_0^R\int_{V\cap S^2}\frac{1}{r^{d-1}\Vert\nabla g({\bf x})\Vert}\,r d{\bf x}dr\approx 2a \frac{R^{3-d}}{3-d}\int_{V\cap S^2}\frac{1}{\Vert\nabla g({\bf x})\Vert}\,d{\bf x}$$for $d\not=3$(I'm ignoring the part of the integral near the origin in $\mathbb{R}^3$, as $\nabla g=0$ there, and the estimate of the volume becomes very inaccurate). Multiplying by $\zeta(3)^{-1}$ gives the "expected number" of coprime integer solutions $\Vert{\bf x}\Vert\le R$. Note that this tends to infinity as $R\to\infty$ for $d < 3$. If $d=3$, then we should replace the $R^{3-d}/(3-d)$ term by $\log R$, so it still tends to infinity. On the other hand, for $d\ge4$, we have a finite volume. So, we should expect the number of coprime solutions ${\bf x}\in\mathbb{Z}^3$ with $\vert g({\bf x})\vert\le a$ to be finite when $d\ge4$ and probably infinite when $d\le3$.
That's only part of the story though. We should also look at the probability that $g({\bf x})=0$ is satisfied modulo prime powers $p^r$. If this probability is about $p^{-r}$, as you might expect, then the heuristic argument above seems good. If the probability is much more than $p^{-r}$ (as is the case if we don't restrict to coprime triples), then the argument above only gives a lower estimate and, if the probability is much less than $p^{-r}$, then we expect fewer solutions than the argument above would suggest.