In light of the fact that I've made multiple mistakes whilst thinking this through, I will try to write a somewhat thorough answer, so bear with me. Let $S^1\rightarrow M\stackrel{\pi}{\rightarrow}B$ be an orientable circle bundle whose base is a path-connected CW-complex. (I am assuming this for the sake of simplicity, since you seem to care for smooth manifolds. The results can be generalized to numerable bundles whose base has the homotopy type of a CW-complex.) Fix basepoints throughout, though I will omit them from notation, and a universal cover $p\colon\tilde{B}\rightarrow B$ of $B$. I claim that $\pi_1(M)$ is finite if and only if $\pi_1(B)$ is finite and the pullback $S^1\rightarrow p^{\ast}(M)\stackrel{p^{\ast}\pi}{\rightarrow}\tilde{B}$ is a non-trivial bundle. This is the case if and only if $\pi_1(B)$ is finite and the Euler class $u({\pi})\in H^2(B)$ is not in the kernel of $p^{\ast}\colon H^2(B)\rightarrow H^2(\tilde{B})$. Two notable corollaries:
If $S^1\rightarrow M\stackrel{\pi}{\rightarrow}B$ is a non-trivial, orientable circle bundle over a simply connected CW-complex, then $\pi_1(M)$ is finite.
If $S^1\rightarrow M\stackrel{\pi}{\rightarrow}B$ is an orientable circle bundle over a path-connected CW-complex and $u(\pi)\in H^2(B)$ is torsion, then $\pi_1(M)$ is infinite.
The orientability assumption says that the bundle has structure group $\mathrm{Homeo}^+(S^1)$, the orientation-preserving homeomorphisms of $S^1$, hence is classified by a map (unique up to homotopy) $B\rightarrow B\mathrm{Homeo}_+(S^1)$. Note that $\mathrm{Homeo}^+(S^1)\cong\mathrm{Homeo}^+_e(S^1)\times S^1$, where $\mathrm{Homeo}_e^+(S^1)$ is the group of orientation-preserving homeomorphisms of $S^1$ fixing the identity. This group is isomorphic to the group $\mathrm{Homeo}_{\partial I}(I)$ of homeomorphisms of $I$ that fix the boundary point-wise, which is contractible by a straight-line homotopy. Thus, $B\mathrm{Homeo}^+(S^1)\simeq B\mathrm{Homeo}^+_e(S^1)\times BS^1\simeq BS^1\simeq\mathbb{CP}^{\infty}$ (the last homotopy equivalence due to the explicit model $S^1\rightarrow S^{\infty}\rightarrow\mathbb{CP}^{\infty}$ of a universal principal $S^1$-bundle). However, $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z},2)$, so ultimately the bundle is classified by an element of $[B,B\mathrm{Homeo}^+(S^1)]=[B,\mathbb{CP}^{\infty}]=H^2(B)$, a characteristic class $u(\pi)$ called the Euler class. More precisely, the universal element $u\in H^2(\mathbb{CP}^{\infty})$ that classifies the universal bundle is a generator and if $S^1\rightarrow M\stackrel{\pi}{\rightarrow}B$ is classified by $f\colon B\rightarrow\mathbb{CP}^{\infty}$, then $u(\pi)=f^{\ast}(u)$.
To study $\pi_1(M)$, our primary tool is the LES of a fibration, which reads
$$\require{AMScd}
\begin{CD}
\dotsc @>>> \pi_2(B) @>{\partial}>> \pi_1(S^1) @>>> \pi_1(M) @>>> \pi_1(B) @>>> 1.
\end{CD}$$
Thus, $\pi_1(M)$ is an extension of $\pi_1(B)$ by $\mathrm{coker}(\pi_2(B)\rightarrow\pi_1(S^1))$, hence $\pi_1(M)$ is finite if and only if $\pi_1(B)$ and $\mathrm{coker}(\pi_2(B)\rightarrow\pi_1(S^1))$ are finite. The group $\pi_1(S^1)\cong\mathbb{Z}$ is infinite, yet all of its non-trivial quotients are finite. Thus, $\pi_1(M)$ is finite if and only if $\pi_1(B)$ is finite and $\partial\colon\pi_2(B)\rightarrow\pi_1(S^1)$ is non-trivial. This latter condition can be analyzed further. To do this, let me set up two commutative diagrams, namely
$$\require{AMScd}
\begin{CD}
\dotsc @>>> \pi_2(\mathbb{CP}^{\infty}) @>{\partial}>> \pi_1(S^1) @>>> \pi_1(\mathbb{C}^{\infty})=1 @>>> \pi_1(\mathbb{CP}^{\infty})=1 @>>> 1\\
{} @AAA @A{\mathrm{id}}AA @AAA @AAA\\
\dotsc @>>> \pi_2(B) @>{\partial}>> \pi_1(S^1) @>>> \pi_1(M) @>>> \pi_1(B) @>>> 1\\
{} @A{\wr}AA @A{\mathrm{id}}AA @AAA @AAA\\
\dotsc @>>> \pi_2(\tilde{B}) @>{\partial}>> \pi_1(S^1) @>>> \pi_1(p^{\ast}M) @>>> \pi_1(\tilde{B})=1 @>>> 1
\end{CD}$$
and
$$\require{AMScd}
\begin{CD}
H^2(\tilde{B}) @>{\sim}>> \mathrm{Hom}(H_2(\tilde{B}),\mathbb{Z}) @>{\sim}>> \mathrm{Hom}(\pi_2(\tilde{B}),\mathbb{Z})\\
@AAA @AAA @A{\wr}AA\\
H^2(B) @>>> \mathrm{Hom}(H_2(B),\mathbb{Z}) @>>> \mathrm{Hom}(\pi_2(B),\mathbb{Z})\\
@AAA @AAA @AAA\\
H^2(\mathbb{CP}^{\infty}) @>{\sim}>> \mathrm{Hom}(H_2(\mathbb{CP}^{\infty}),\mathbb{Z}) @>{\sim}>> \mathrm{Hom}(\pi_2(\mathbb{CP}^{\infty}),\mathbb{Z}).
\end{CD}$$
They are both induced by the universal covering $\tilde{B}\rightarrow B$ and the classifying map $B\rightarrow\mathbb{CP}^{\infty}$. The former is the naturality diagram comparing the LESs of the respective bundles. The latter is the naturality diagram comparing the respective Hurewicz maps. The horizontal isomorphisms are due to the universal coefficient and Hurewicz theorems, since $\tilde{B}$ and $\mathbb{CP}^{\infty}$ are simply connected. The vertical isomorphisms are due to the fact that a universal cover induces isomorphisms on higher homotopy groups.
Identifying $\pi_1(S^1)=\mathbb{Z}$, the boundary maps in the respective LESs in diagram $1$ are identified with elements of the $\mathrm{Hom}$-groups in the right column of diagram $2$. The commutativity of diagram $1$ implies that these elements map to one another under the vertical arrows in the right column of diagram $2$. The boundary map of the universal bundle is an isomorphism (to its left comes $\pi_2(\mathbb{C}^{\infty})=1$), so it generates $\mathrm{Hom}(\pi_2(\mathbb{CP}^{\infty}),\mathbb{Z})$ and the isomorphisms in the bottom row of diagram $2$ imply that it is, up to sign depending on the chosen generators, the image of the universal element $u\in H^2(\mathbb{CP}^{\infty})$. The lower part of diagram $2$ then implies that the Euler class $u(\pi)\in H^2(B)$ maps to the boundary map $\partial\in\mathrm{Hom}(\pi_2(B),\mathbb{Z})$, perhaps up to sign. The isomorphisms in the upper part of diagram $2$ then imply that this element is trivial if and only if $u(\pi)$ is in the kernel of $p^{\ast}\colon H^2(B)\rightarrow H^2(\tilde{B})$. However, naturality implies that $p^{\ast}(u(\pi))=u(p^{\ast}\pi)$, so this is the case if and only if the pullback bundle $p^{\ast}\pi\colon p^{\ast}M\rightarrow\tilde{B}$ is trivial.
Edit: I claim furthermore that, if $\pi_1(B)$ is finite, the kernel of $H^2(B)\rightarrow H^2(\tilde{B})$ is precisely the torsion subgroup $TH^2(B)$ of $H^2(B)$. To see this, consider the natural diagram of SESs given by the universal coefficient theorem,
$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> H^2(\tilde{B}) @>{\sim}>> \mathrm{Hom}(H_2(\tilde{B}),\mathbb{Z}) @>>> 0\\
{} @AAA @AAA @AAA\\
0 @>>> \mathrm{Ext}(H_1(B),\mathbb{Z}) @>>> H^2(B) @>>> \mathrm{Hom}(H_2(B),\mathbb{Z}) @>>> 0.
\end{CD}$$
Since $H_1(B)=\pi_1(B)^{\mathrm{ab}}$ is finite, $\mathrm{Ext}(H_1(B),\mathbb{Z})$ is torsion. Evidently, $\mathrm{Hom}(H_2(B),\mathbb{Z})$ is torsion-free, so it follows from exactness that $\mathrm{Ext}(H_1(B),\mathbb{Z})$ is the torsion subgroup of $H^2(B)$. The claim is now, by a short diagram chase, equivalent to the injectivity of $\mathrm{Hom}(H_2(B),\mathbb{Z})\rightarrow\mathrm{Hom}(H_2(\tilde{B}),\mathbb{Z})$. This map is dual to the map $H_2(\tilde{B})\rightarrow H_2(B)$, which fits into the commutative square
$$\require{AMScd}
\begin{CD}
\pi_2(\tilde{B}) @>{\sim}>> H_2(\tilde{B})\\
@V{\wr}VV @VVV\\
\pi_2(B) @>>> H_2(B).
\end{CD}$$
The horizontal maps are Hurewicz homomorphisms. Thus, the cokernel $C$ of $H_2(\tilde{B})\rightarrow H_2(B)$ is the cokernel of $\pi_2(B)\rightarrow H_2(B)$, which is isomorphic to $H_2(K(\pi_1(B),1))$ (see Hatcher, p.390/391). This group is finite since $\pi_1(B)$ is finite (it is finitely generated since $K(\pi_1(B),1)$ can be realized as a simplicial complex with finitely many simplices in each dimension, so its homology is computed by a chain complex of finitely generated abelian groups, and it has no free summand cause the rational homology vanishes, because the rational homology of its universal cover vanishes (see e.g. here) as that universal cover is contractible). Now, applying the left-exact functor $\mathrm{Hom}(-,\mathbb{Z})$ transforms the half-exact
sequence $H_2(\tilde{B})\rightarrow H_2(B)\rightarrow C\rightarrow 0$ into the half-exact sequence $0\rightarrow\mathrm{Hom}(C,\mathbb{Z})=0\rightarrow\mathrm{Hom}(H_2(B),\mathbb{Z})\rightarrow\mathrm{Hom}(H_2(\tilde{B}),\mathbb{Z})$. Thus, the map is injective, as desired.
