Groups that are not amalgams It is clear that cyclic groups have the property that they cannot be written in a non-trivial way as an amalgamated free product or as an HNN-extension.
Can someone please provide us examples of torsion-free 2-generated groups having this property? Any comments related to this question are welcome!
[I'm aware of Serre's Theorem 15 in Trees.]

Edit: The term "in a non-trivial way" means the following.
For an amalgamated free product $G=H*_C K$ it is required $H\ne C\ne K$. For an HNN-extension $G=HNN(H,A,B,t)$ it is required $A\ne H\ne B$.
 A: Actually, your first statement is false for the infinite cyclic group:
$\mathbb{Z}=1*_1$
the trivial HNN extension of the trivial group.  The point is that, by the ideas described in Trees, a splitting of a group is the same thing as an action of that group on a tree, and there is an obvious action of $\mathbb{Z}$ on the real line, which is a tree.
A group that does not split is said to have Property FA.  There are many constructions of such groups, though some criteria involve torsion, so your torsion-free restriction makes the question more interesting.
One source of examples is given by random groups, which can be two-generated.  Dahmani, Guirardel and Przytycki showed that random groups have Property FA.
ADDED:
The following is missing some details; the gaps can be filled in from Serre's Trees and from Peter Scott's article `The Geometries of 3-manifolds'.  I'll fill in some more details if you ask for them.
Here's an explicit example.  You probably know that triangle groups
$T(p,q,r)\cong \langle a,b\mid a^p=b^q=(ab)^r=1)\rangle$
have Property FA. This is quite  easy to see by thinking about actions on trees.  Because they are all of finite order, $a,b$ and $ab$ each have to fix vertices, and the only way this can happen is if they all fix the same vertex.  These groups are 2-generator, but clearly not torsion-free.
Now here's a torsion-free example, which is the fundamental group of a certain Seifert-fibred space.  For details on Seifert-fibred spaces, I recommend Peter Scott's article `The Geometries of 3-manifolds'.
Let $\mathrm{gcd}(p,p')=\mathrm{gcd}(q,q')=\mathrm{gcd}(r,r')=1$. The group in question is
$\Gamma=\langle a,b,z\mid a^p=z^{p'}, b^q=z^{q'}, (ab)^r=z^{r'}, [a,z]=[b,z]=1\rangle~.$
Note that if $\mathrm{gcd}(p',q',r')=1$ then $\Gamma$ is 2-generated; $\Gamma$ is also torsion-free. $\Gamma$ can be written as a non-split central extension
$1\to\langle z\rangle\to \Gamma\to T(p,q,r)\to 1~.$
Suppose that $\Gamma$ acts on a tree $T$.  Then, using some standard lemmas about actions on trees and the fact that it is normal, either $Z=\langle z\rangle$  fixes $T$ pointwise (in which case we are done), or $T$ can be taken to be a line which $z$ translates.  
In this latter case, translation distance gives a homomorphism $f:\Gamma\to\mathbb{Z}$ with $f(z)$ non-trivial.  But the existence of this homomorphism implies that
$p'/p+q'/q=r'/r$
which of course won't be true for most choices of parameters.
