Generalizing a property of the two sphere There are many circle bundles over the sphere $ S^2 $ (in fact infinitely many) but all of them are principal.
Do there exist any other manifolds besides $ S^2 $ for which (nontrivial bundles exist and) all circle bundle are principal?
I believe that $ S^2 $ is the only surface with this property.
For example $ \mathbb{R}P^2 $ admits infinitely many non principal circle bundles, as does the Klein bottle $ K^2 $. And the torus $ T^2 $ admits two different non principal circle bundles. And the plane $ \mathbb{R}^2 $ is contractible so all circle bundles are principal but also all circle bundles are trivial (indeed it is true for all $ n $ that all circle bundles over $ \mathbb{R}^n $ are trivial, in fact even all fiber bundles are trivial see A fiber bundle over Euclidean space is trivial.).
Also note that the Klein bottle $ K^2 $ is a non principal circle bundle over $ S^1 $. So the circle does not have this property.
Higher spheres $ S^n $ for $ n \geq 3 $ are other examples where this property fails, since every circle bundle is trivial, this follows from the fact that $ \pi_{n-1}(S^1)=0 $ for $ n\geq 3 $ see for example https://math.stackexchange.com/a/269761/758507.
 A: First, you didn't state the structure group of the bundle, but $Diff(S^1)$ deformation retracs to $O(2)\subseteq Diff(S^1)$, so we can assume that you mean $O(2)$.  Said another way, all $S^1$-bundles over a manifold are the unit circle bundles in a rank $2$ vector bundle.
Also, for ease of writing, I'll write $G$ for the group $\mathbb{Z}/2\mathbb{Z}$.
Claim 1:  For a manifold $M$, all circle bundles over $M$ are principal iff $H^1(M;G) = 0$.
Proof:  Assume first that $H^1(M;G)$ is trivial.  And choose a circle bundle $E\rightarrow M$.  As mentioned above, we may assume the bundle is linear, so we can talk about characteristic classes.  By assumption, $w_1(E)\in H^1(M;G) = 0$, so the bundle is orientable.  This mean the structure group is actually $Diff^+(S^1)$, which deformation retracts to $SO(2)$.  But this, then, means the bundle is principal.
Conversely, assume that $0\neq x \in H^1(M;G)$.  Since $\mathbb{R}P^\infty$ is a $K(G,1)$, there must be a map $\phi:M\rightarrow \mathbb{R}P^\infty$ for which $\phi^\ast:H^1(\mathbb{R}P^\infty;G)\rightarrow H^1(M;G)$ has image $x$.  Pulling back the tautological real rank $1$ bundle over $\mathbb{R}P^\infty$ along $\phi$, we obtain a rank $1$ vector bundle $E$ over $M$ with $w_1(E) = x\neq 0$.
Now, consider the vector bundle $E\oplus 1$ where $1$ denotes a trivial rank $1$-bundle.  By the Whitney sum formula, $w_1(E\oplus 1) = w_1(E) = x$.  Finally, the unit sphere bundle in $E\oplus 1$ gives a circle bundle over $M$ with $w_1$ non-trivial.  Hence this bundle is not orientable and thus not principal.  $\square$.
From claim 1, it follows that among closed surfaces, $S^2$ is the only one for which all circle bundles are principal.
Claim 2:  For a manifold $M$, there is a non-trivial circle bundle over $M$ iff $H^2(M;\mathbb{Z})\neq 0$.
Proof:  Such bundles are classified by their Euler class, an element of $H^2(M)$.  In more detail, any element $x\in H^2(M)$ corresponds to a map to $K(\mathbb{Z},2) = \mathbb{C}P^\infty$.  Pulling back the tautological rank 1 complex vector bundle over $\mathbb{C}P^\infty$ gives a circle bundle over $M$ with Euler class $x$.  Moreover, all principal $S^1$ bundles arise from this construction since $BS^1 = \mathbb{C}P^\infty$.  $\square$
Combining claim 1 and claim 2, $M$ meets your requirements iff $H^1(M;G) = 0$ and $H^2(M;\mathbb{Z})\neq 0$.
Now getting examples is easy.  The following are all examples of what you want:

*

*If $H^1(N;G) = 0$, then $N\times S^2$ is an example.  More generally, if $M$ is any example of what you want, then $M\times N$ is an example.

*$\mathbb{C}P^n$ for every $n$

*Any non-abelian connected compact Lie group mod its maximal torus.

*Any simply connected closed $4$-manifold except $S^4$.

*If $M$ has the property and $H^1(N;G) = 0$, and $\dim M = \dim N \geq 4$, then $M\sharp N$ has the property.

*Any lens space $L_{p,q}$ with $p$ odd.

