Over which fields does the Mordell-Weil theorem hold? According to a well-known theorem of Mordell, the group of rational points $E(\mathbf{Q})$ of an elliptic curve $E/\mathbf{Q}$ is finitely generated. Weil generalized this theorem to abelian varieties over number fields.
Less well-known is the following generalization, due I believe to Néron : if $K$ is a field of finite type (that is, finitely generated over its prime field) and $A$ is an abelian variety over $K$, then $A(K)$ is finitely generated. There is an even more general statement, the Lang-Néron theorem, for relative field extensions which are finitely generated (see Brian Conrad's article for the precise statement and a proof of this theorem).
Q1. Are there other fields $K$ for which the group of $K$-rational points of an abelian variety over $K$ is always finitely generated?
In the other direction, there exist fields $K$ for which $A(K)$ is clearly never finitely generated whenever $\operatorname{dim}(A) \geq 1$. For example $K=\mathbf{C}$, in which case it follows from the description af abelian varieties as complex tori. If $K$ is a finite extension of $\mathbf{Q}_p$, then $A(K)$ contains a finite-index subgroup isomorphic to $\mathcal{O}_K^{\operatorname{dim} A}$, so $A(K)$ is again never finitely generated. Other examples I can think of are complete discretely valued fields and algebraically closed fields. Note that we often have the stronger result that $A(K) \otimes \mathbf{Q}$ is infinite-dimensional (except when $K=\overline{\mathbf{F}}_p$, in which case $A(K)$ is a torsion group).
Q2. Are there other fields $K$ for which the group of $K$-rational points of a non-trivial abelian variety over $K$ is never finitely generated?
 A: This is a partial answer to Q2, in particular a full answer to the question asked by François Brunault in the comments:

By the way, if K is ferile, will A(K)⊗Q be necessarily be infinite-dimensional? 

Here, A is a non-trivial abelian variety over K, and fertile = large = ample = for every geometrically integral smooth K-variety the set of K-rational points is either empty or Zariski-dense.
If K is algebraic over a finite field, the answer is NO, as already pointed out by Laurent Moret-Bailly. In all other cases, the answer is YES: 
For fields of infinite transcendence degree, this was proven in the paper On the rank of abelian varieties over ample fields with Sebastian Petersen using elementary methods.
For fields of characteristic zero, in the same paper we deduce the claim using the Mordell-Lang conjecture as proven by Faltings, Raynaud and possibly others.
For fields of positive characteristic and transcendence degree at least one, in Ranks of abelian varieties and the full Mordell-Lang conjecture in dimension one we deduce the claim from a recent result of Rössler by a similar approach. 
A: This is an attempt at a relatively mild generalization of what others have said:
Let $K$ be a field and $|\cdot|: K \rightarrow \mathbb{R}$ be a nontrivial absolute value on $K$. 
$\bullet$ If $K$ is complete for $|\cdot|$, then $E(K)$ has the structure of a $K$-analytic Lie group in the sense of Serre.  In particular it is a $K$-analytic manifold 
so has at least continuum cardinality.
$\bullet$ When $|\cdot|$ comes from a rank one valuation $v$, I suspect that even if $K$ is merely Henselian for $v$, then $E(K)$ cannot be finitely generated. 
Here is a proof in the case that the valuation is discrete and the residue field $k$ is infinite: standard arguments involving the formal group still give a filtration 
$E(K) \supset E^0(K) \supset E^1(K) \supset E^2(K) \supset \ldots$ 
such that (by Hensel's Lemma) for all $n \geq 1$, $E^n(K)/E^{n+1}(K) \cong (k,+)$.  (Just last night I noticed that Cassels's Lectures on Elliptic Curves has a beautiful, elementary take on this.  He works with the case $K = \mathbb{Q}_p$ but the argument holds much more generally.)  If $k$ is infinite, then its additive group is not finitely generated and thus $E(K)$, having a subquotient which is not finitely generated, is itself not finitely generated.
A: Here is an [INCOMPLETE, POSSIBLY INCORRECT] answer to question 1. Yes. Let $C_n/k,n=1,2,\ldots$ be a sequence of curves of increasing genus defined over a finite field $k$ with maps $C_{n+1} \to C_n$ for all $n$. Let $K = \bigcup k(C_n)$. Assume further that $Jac(C_{n+1})/Jac(C_n)$ is simple for all $n$, where $Jac$ is the Jacobian (this is will be the typical case in such a tower). Then $A(K)$ is finitely generated for any abelian variety $A$, as $A(K) = A(k(C_n))$ where $n$ is the largest integer for which $A$ occurs as a factor of $Jac(C_n)$. 
EDIT: As pointed out by Will in the comment below, this only works if $A$ is defined over $k$. 
Here is an answer for question 2. Yes. Let $K$ be an infinite subfield of the algebraic closure of a finite field. It follows easily from the Weil bound that $A(K)$ is an infinite torsion group so is not finitely generated. 
A: [Edited November 2 for brevity]  
(1) An extension of the finitely generated case for Q1: let $K_0$ be finitely generated over the prime field, and let $K=K_0((x_i)_{i\in I})$ be a purely transcendental extension of $K_0$. Then $K$ "satisfies Q1". Indeed, any abelian variety $A/K$ is defined over some intermediate $K_1:=K_0((x_i)_{i\in J})$, $J\subset I$ finite. Then $A(K_1)$ is finitely generated, but $A(K)=A(K_1)$ since $K/K_1$ is purely transcendental.  
(2) Another "easy" case for Q2: if $K/\mathbb{F}_p$ is infinite algebraic, then for any $A$ the group $A(K)$ is torsion, but must be infinite by Weil's estimates, hence is not finitely generated.  
(3) A general result on Q2:  Say a field $K$ is fertile if for every smooth irreducible $K$-variety $X$, if $X(K)$ is nonempty, then it is Zariski-dense.
(Pop, who invented the concept, called these fields "large"; others say "ample").  
I claim that every fertile field $K$ satisfies Q2.  This includes in particular:  
(3a) all Henselian valued fields (already mentioned by Pete, but there is no restriction on the rank here, except the valuation must be nontrivial). 
(3b) Pseudo-algebraically closed fields (i.e. such that every geometrically irreducible variety has a rational point). This includes example (2) above.  
Proof of claim: Let $A$ be an abelian $K$-variety of dimension $g>0$, with origin $e$. We may assume $g\geq2$ (if $g=1$, consider $A\times A$). Let $t_1,\dots,t_g$ be a regular system of parameters at $e$. Consider the rational map $(t_1:\dots:t_g):A\dots\to\mathbb{P}^{g-1}_K$. It induces a morphism $f:U\smallsetminus\{e\}\to\mathbb{P}^{g-1}_K$ where $U\subset A$ is a neighborhood of $e$. Let $\widetilde{U}$ be the blow-up of $e$ in $U$. By the assumption on $t_1,\dots,t_g$, we get a morphism $\widetilde{f}:\widetilde{U}\to \mathbb{P}^{g-1}_K$ which induces an isomorphism  $E\to\mathbb{P}^{g-1}_K$where  $E$ is the exceptional divisor. Moreover, $\widetilde{f}$ is smooth along $E$. Shrinking $U$, we may assume $\widetilde{f}$ smooth.
For every $y\in\mathbb{P}^{g-1}(K)$, $\widetilde{f}^{-1}(y)$ is a smooth curve with a rational point on $E$. Since $K$ is fertile, $\widetilde{f}^{-1}(y)$  also has rational points on $U\smallsetminus\{e\}$. Hence $f:U(K)\smallsetminus\{e\}\to\mathbb{P}^{g-1}(K)$ is surjective.
On the other hand, if $A(K)$ were finitely generated there would be a finitely generated subfield of definition $K_0\subset K$ for $A$, $U$ and $f$ such that $A(K)=A(K_0)$, which would imply $f(U(K))\subset\mathbb{P}^{g-1}(K_0)$. This is a contradiction because $K_0\neq K$ (finitely generated fields are not fertile).
