Einstein Field Equations in other space-time dimensions than 3+1? This question is apparently quite simple but I can't seem to find an answer to it, so I was hopping anyone could clarify me.
Are the Einstein field equations (EFE) only valid for a 3+1 dimensional space-time?
I've read somewhere, which I can't remember or find, that there were problems with the EFE in a 2+1 dimensions...Why would that be? 
What about 1+1?
 A: There is nothing "wrong" with the Einstein field equations in $2+1$ as indicated by the comments, but they do have interesting, significantly restricted behavior in $2+1$ dimensions.
For example, the Wikipedia page referred to by Olof in the comments says that in $2+1$, every vacuum solution is locally either $\mathbb R^{2,1}$, $\mathrm{AdS_3}$, or $\mathrm{dS}_3$.  Here's why.  In $d+1$ with $d\neq 1$, the vacuum field equations (those with $T_{\mu\nu} =0$) can be manipulated to show that
$$
  R_{\mu\nu} = \frac{R}{d+1}g_{\mu\nu}
$$
On the other hand, one can show (see Weinberg Gravitation and Cosmology eq. 6.7.6) that in $2+1$, the Riemann tensor satisfies
$$
R_{\lambda\mu\nu\kappa}
    = g_{\lambda\nu} R_{\mu\kappa} - g_{\lambda\kappa}R_{\mu\nu} - g_{\mu\nu}R_{\lambda\kappa} + g_{\mu\kappa} R_{\lambda\nu} - \frac{1}{2}(g_{\lambda\nu}g_{\mu\kappa} - g_{\lambda\kappa}g_{\mu\nu})R
$$
and combining these results gives
$$
  R_{\lambda\mu\nu\kappa}
    = \frac{1}{6}(g_{\lambda\nu}g_{\mu\kappa}-g_{\lambda\kappa}g_{\mu\nu})R
$$
which is precisely the Riemann tensor for a maximally symmetric spacetime in $2+1$ which gives the result.
Notice that this behavior is in stark contrast to the vacuum behavior in $3+1$.  For example, take the vacuum region outside of a spherically symmetric massive body in $3+1$ (like a black hole).  This region is not flat, but in $2+1$ with vanishing cosmological constant any vacuum region outside of a massive body would be.  Pretty strange.
A: Joshphysics has already given a nice answer showing that in 2+1 dimensional Einstein gravity any metric is locally equivalent to a metric of constant curvature. As dilaton mentioned in a comment this in particular means that there are no local excitations.
The updated question also asks about 1+1 dimensions. In this case the answer is even simpler: the 1+1 dimensional Einstein tensor $R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$ vanishes identically. Hence the Einstein-Hilbert action $\int d^2x \sqrt{-g}R$ is a surface term.
