Why should I believe the Mordell Conjecture?
Here's a quick and dirty version of George Lowther's calculation that I learned from Bjorn Poonen. It is presented in a bit more generality in the next-to-last slide of this talk, so it is in a sense pre-masticated for colloquium use:
Let $B$ be a large positive integer which we will allow to grow. The set of integer points in the box $[-B,B]^{\times 3}$ has size of order $B^3$. A homogeneous degree $d$ polynomial in 3 variables with integer coefficients (describing a degree $d$ plane curve over $\mathbb{Q}$) will take values of size about $B^d$ when inputs are taken from the box. If we assume the values are uniformly distributed, the expected number of zeroes in the box is $B^{3-d}$.
There is a qualitative difference in expected number of zeroes based on the sign of the exponent $3-d$. This suggests that plane curves of degree more than 3 (namely those of genus at least 2) will only have "accidental" rational points.
If the curve $X$ (over the number field $k$) has no $k$-points at all, then Mordell's conjecture is true for $X$. Otherwise, if $O$ is a given $k$-point on $X$, we can get a map (the Albanese map) $X \to Jac(X)$ via $P \mapsto P - O$, embedding $X$ as a $k$-subvariety of $Jac(X)$.
The Mordell--Weil theorem shows that $Jac(X)(k)$ is a finitely generated abelian group. Since $g \geq 2$ by assumption, the curve $X$ is of positive codimension in $Jac(X)$, and so it is not unreasonable to imagine that $X$ intersects $Jac(X)(k)$ in only finitely many points. (George Lowther's answer, posted while I was writing this, describes the same heuristic.)
In fact, my understanding is that Weil proved the general form of the Mordell--Weil theorem (in his thesis, I think) precisely to try to implement this strategy and so prove Mordell's conjecture. Unfortunately, no-one has been able to implement this strategy quite as cleanly as the intuition above suggests, although it inspired an important approach, due to Chabauty, which established the Mordell conjecture in some non-trivial cases; see this paper by McCallum and Poonen for an explanation of Chabauty's method, and Coleman's strengthening of it.
More recently, Minhyong Kim has developed an anabelian strengthening of Chabauty's method; see e.g. here and here. The second paper (joint with Coates) gives a new proof of Mordell for Fermat curve's, among other examples.
Disclaimer In the spirit of the question I am not trying to be very precise below. The key word is motivation. :)
Reason #1 (simple, but lame)
Fermat's Last Theorem says that a curve of that kind of a particular form has no rational points. One could consider Mordell's conjecture as a weak form of that. This is probably not very convincing but perhaps a good way to start a colloquium. A derivative of this idea is that you could look at a lot of examples and make this conjecture.
Reason #2 (hyperbolicity and rational curves)
Another, perhaps more convincing idea is that there is a trichotomy of curves given by $g=0$, $g=1$, and $g\geq 2$. If you look at topological, geometric, arithmetic properties of these curves, their properties align very strongly with these classes.
Having genus at least $2$ in other words means that the curve is of general type. From a differential geometric point of view that means (mostly) negative curvature or (roughly) being hyperbolic. A compact complex analytic space is hyperbolic if and only if there are no non-constant holomorphic maps to the space from $\mathbb C$. Algebraically this roughly means that there are no rational curves or abelian varieties in the space. This is of course not entirely true, but we have
Lang's Conjecture Let $X$ be a variety of general type, then there exists a proper closed subvariety $Z\subsetneq X$ such that any rational curve or abelian variety contained in $X$ is contained in $Z$.
The existence of rational curves in the geometric setting is very closely related to the existence of rational points in the arithmetic setting. So, it is a natural idea that if a space has properties that suggest that it does not contain many rational curves then if it is defined over a number field then it should not contain many rational points.
In fact there is an analogue of Lang's conjecture for number fields and rational points which reduces to the Mordell conjecture for curves:
Lang's conjecture (number field version) Let $X$ be a variety of general type defined over a number field $F$. Then the set of $F$-rational points of $X$ is not dense in $X$.
Then at this point one (who is more qualified to do this than I) could explain how this is taken further by Vojta (see Matt's comment below). Then one gets all kinds of interesting conjectures estimating various heights. The motivation for these come from Nevanlinna theory and analogies between results on value distribution of meromorphic functions and Diophantine approximation.
Reason #3 (Shafarevich's Conjecture, de Franchis's Theorem)
Faltings proved the Mordell Conjecture using Parshin's trick and proving Shafarevich's Conjecture. These are the following:
Shafarevich's Conjecture Let $F$ be either a number field or the function field of a curve defined over an algebraically closed field of characteristic $0$. Let $R\subset F$ be a subring; if $F$ is a number field, let $R$ be the ring of integers, if $F$ is a function field, let $R$ be the coordinate ring of a smooth affine curve whose fraction field is $F$. Let $\Delta\subset {\rm Spec} R$ be a finite set of primes. A smooth projective curve over $F$ is called admissible (with respect to $\Delta$) if it has good reduction outside $\Delta$ and in the case $F$ is a function field it is also non-isotrivial. Then the Shafarevich conjecture is that for a fixed $F$ and $\Delta$ there exist only finitely many admissible curves of a fixed genus$\geq 2$.
Parshin's Trick Shafarevich's conjecture implies Mordell's Conjecture.
The idea of Parshin's trick: Mordell's conjecture says that if $X$ is a smooth projective curve over $F$ of genus at least $2$ then there are only finitely many sections of the structure map $X\to {\rm Spec F}$. Parshin's idea is that for any such section one can obtain a new curve by a cover that's ramified exactly over the section. One proves that this can be done so the genus of the resulting curve is bounded in terms of the initial data.
This way one associates a set of curves with bounded genus to the set of sections of a fixed curve and concludes that Shafarevich's conjecture implies Mordell's.
Truth be told, one also needs to prove that this association is at worst finite-to-one in the sense that there cannot be infinitely many sections producing the same (abstract) covering curve. But this follows from de Franchis's (classical) theorem:
de Franchis's Theorem Let $C$ and $D$ be smooth curves of genus at least $2$. Then there exist only finitely many dominant rational maps $D\to C$.
So, at this point you might say: OK, but why should Shafarevich's conjecture be true?
Let's try to answer that. I will concentrate on the function field case for two reason. Mostly because that is what I know a little about and second because I claim that for motivation it is good enough if you know why it should be true in that case. The number field case ought to behave similarly.
What does Shafarevich's conjecture mean in terms of moduli? One could say that the function field version is equivalent to the following:
Shafarevich's Conjecture 2.0 (function field case) Let $B$ be a smooth curve. Then there exist only finitely many non-trivial maps $B\to \mathscr M_g$ where $\mathscr M_g$ is the moduli stack of smooth projective curves of genus $g$.
But this is "just" a higher dimensional log version of de Franchis's Theorem: $B$ and ${\mathscr M}_g$ are of log general type, so we expect only finitely many non-trivial maps between them (remembering that $\dim B=1$).
Remark I think this is a pretty good reason to believe Mordell's conjecture now. Obviously this was not known to those who believed it before the (late) sixties.