The ring $A_{\text{inf}}$ is local, so every vector bundle (i.e. finite projective module, or finitely presented flat module) is free [Stacks, Tag 00NX(4)].

Now let $(R,\mathfrak m)$ be any local ring, and let $X = \operatorname{Spec} R$, and $U = X \setminus\{\mathfrak m\}$. Assume that the natural map $R \to \Gamma(U,\mathcal O_U)$ is an isomorphism.

To show that the restriction is fully faithful, let $\mathscr E$ and $\mathscr F$ be vector bundles on $X$. By the above, we have $\mathscr E \cong \mathcal O_X^e$ and $\mathscr F \cong \mathcal O_X^f$ for some $e, f$. Then $\operatorname{Hom}_X(\mathscr E,\mathscr F) \cong \mathcal O_X^{ef}$, so the natural map
$$\begin{array}{ccc}\operatorname{Hom}_X(\mathscr E,\mathscr F) & \to & \operatorname{Hom}_U(\mathscr E|_U,\mathscr F|_U) \\
|| & & ||\\
\Gamma(X,\mathcal O_X^{ef}) & \to & \Gamma(U,\mathcal O_U^{ef})\end{array}$$
is an isomorphism since $R \to \Gamma(U,\mathcal O_U)$ is. $\square$

In turn, assume $\mathfrak m = (x,y)$ for some elements $x, y \in R$. Set $R_1 = R[1/x]$, $R_2 = R[1/y]$, and $R_{12} = R[1/xy]$, and let $U_i$ (resp. $U_{ij}$) be $\operatorname{Spec} R_i$ (resp. $\operatorname{Spec} R_{ij}$). Assume the maps $R \to R_i$ and $R_i \to R_{12}$ are injective (e.g. $R$ is a domain).

Now if $R = R_1 \cap R_2 \subseteq R_{12}$, then $R = \Gamma(U,\mathcal O_U)$. Indeed, the sheaf condition on $\mathcal O_X$ gives a short exact sequence
$$\begin{array}{ccccccc}0 & \to & \Gamma(U,\mathcal O_U) & \to & \Gamma(U_1,\mathcal O_{U_1}) \oplus \Gamma(U_2,\mathcal O_{U_2}) & \to & \Gamma(U_{12},\mathcal O_{U_{12}})\\
& & || & & || & & ||\\
0 & \to & \Gamma(U,\mathcal O_U) & \to & R_1 \oplus R_2 & \to & R_{12}, \end{array}$$
realising $\Gamma(U,\mathcal O_U)$ as the intersection $R_1 \cap R_2 \subseteq R_{12}$ (as the second map is $(f,g) \mapsto f-g$). If this equals $R$, then the natural map $R \to \Gamma(U,\mathcal O_U)$ is an isomorphism. $\square$

**References.**

[Stacks] A.J. de Jong et al, *The stacks project*.