For $n\geq 3$, we'll show by induction that if $G$ is a graph with at least $n+1$ vertices in which every set of $n$ vertices have exactly one common neighbour then $G=K_{n+1}$. For $n=3$, take any vertex $v$ of degree $d \geq 3$. Then in $G' = G[\Gamma(v)]$ any two vertices must have exactly one common neighbour and no three vertices can have a common neighbour, which forces $G' = K_3$. So $G=K_4$. For $n > 3$, by induction the subgraph induced by the neighbours of any vertex in $G$ of degree $d\geq n$ is then $K_{n-1}$, completing the proof.