It may be instructive to think about what we really "know" in the case of projective surfaces. Let $k$ be an algebraically closed field of characteristic zero.
Let me first recall some basic definitions.
A projective (integral) variety over $k$ is of general type if some (hence any) desingularization of $X$ has big canonical bundle. Also, a projective variety is pseudo-Mordellic over $k$ if there is a proper closed subset $\Delta\subsetneq X$ such that, for every finitely generated subfield $K\subset k$ and every model $\mathcal{X}$ for $X$ over $K$, the set $\mathcal{X}(K)\setminus \Delta$ is finite.
If you can take $\Delta$ to be the empty set, then we say that $X$ is Mordellic over $k$. A projective variety $X$ is Mordellic over $k$ if and only if every subvariety of $X$ is pseudo-Mordellic over $k$.
Lang's conjecture (in its most optimistic form) is that a projective variety over $k$ is of general type if and only if it is pseudo-Mordellic over $k$
Due to the birational invariance of both the notions of general type and pseudo-Mordellicity, you could as well stick to smooth projective varieties in this conjecture.
In the case of curves, Lang's conjecture boils down to showing that a projective curve is of general type if and only if it is Mordellic which is what Faltings achieved in 1983.
You ask whether Faltings's proof gives maybe a bit more than just the case of curves. Well, it certainly does. Namely, if $X$ is a projective variety over $k$ which admits a (quasi-)finite morphism to the moduli stack of principally polarized abelian varieties (ppav's), then every subvariety of $X$ is of general type (explanation omitted) and $X$ is Mordellic (by Faltings's theorem). Thus, any projective variety admitting a finite morphism to some moduli stack of ppav's will satisfy Lang's conjecture. Examples of such projective varieties are, for example, compact Shimura varieties of abelian type.
What about Faltings's 1992 result on closed subvarieties of abelian varieties? Let us assume that $X$ is a normal projective surface over $k$, and let $q=q(X)$ be the dimension of its Albanese variety. If $q>2$, then you can show that Lang's conjecture holds. That is:
Theorem. (Faltings) Assume that $X$ is a normal projective surface with $q>2$. Then $X$ is of general type if and only if $X$ is pseudo-Mordellic.
Sketch of proof. The fact that a pseudo-Mordellic surface is of general type follows from properties of K3 surfaces and the classification of surfaces; see Remark 7.8 in https://arxiv.org/abs/1909.12187 . Now, let $Y$ be the image of the Albanese map $X\to \mathrm{Alb}(X)$. Note that $Y$ is not an abelian variety, as this would contradict $q>2$. Note that $\dim Y = 1$ or $\dim Y =2$. If $\dim Y = 1$, then $Y$ is a hyperbolic curve (because it is not an abelian variety). The morphism $X\to Y$ is then a fibration of hyperbolic curves over a hyperbolic curve which readily implies that $X$ is pseudo-Mordellic. If $\dim Y=2$, use Faltings's 1991 theorem and Ueno's fibration theorem. QED
Now, this means that in the case of surfaces, it remains to prove Lang's conjecture whenever $q=0$, $q=1$, or $q=2$. However, if $X$ has a finite etale cover $X'\to X$ with $q(X') >2$, then $X'$ satisfies Lang's conjecture by the above. Since Lang's conjecture is "invariant under finite etale covers", it follows that $X$ also satisfies Lang's conjecture. (This is also what Noam Elkies is saying in his answer.) Thus, if $\dim X=2$, Lang's conjecture is really open if the augmented irregularity $\widehat{q}(X) $ is at most two. (The augmented irregularity is the supremum over $q(Y)$ as $Y$ runs over all finite etale covers of $X$.)
This suggests that the next "reasonable" case to study is that of surfaces with augmented irregularity equal to two. (The smaller the augmented irregularity, the further you are from any auxiliary abelian varieties. This makes life considerably harder.)
Using Stein factorization, in the case of surfaces with augmented irregularity two, one is reduced to studying finite ramified covers of abelian varieties. The next step we want to achieve is the following special case of Lang's conjecture:
Conjecture. Let $A$ be an abelian surface over a number field $K$ and let $X\to A$ be a finite ramified covering with $X$ a normal integral surface of general type over $K$. Then $X(K)$ is not dense. (Actually: we expect that $X_{\overline{K}}$ is pseudo-Mordellic, but peu importe.)
This non-density statement is not known.
Usually, Lang's conjecture is interpreted as suggesting that there are not so many points on a variety of general type. However, since $A(K)$ is dense (for some large enough number field $K$), one may also view Lang's conjecture in this specific situation as predicting that there are actually many rational points on an abelian variety. (That is, Lang predicts that there are so many rational points on an abelian variety that even after removing those coming from a ramified covering, the points remain dense.)
This brings me to my final point that some progress was made on this aspect of Lang's conjecture recently; see Theorem 1.3 in https://arxiv.org/abs/2011.12840 . A special case of that result reads as follows:
Theorem. Let $A$ be an abelian variety over a number field $K$ with $A(K)$ dense. Let $\pi:X\to A$ be a finite ramified morphism with $X$ a normal integral variety over $K$. Then $A(K)\setminus \pi(X(K))$ is dense in $A$.
This result says that there are "more" points in $A(K)$ than in $X(K$), as Lang predicted.