I claim that for $k>1$ any such graph is regular, in this case we get a well-known problem (subproblem of describing strongly regular graphs), which does not seem to be solved completely. Case $k=1$ is itself known, friendly vertex exists in this case by the friendship theorem of Paul Erdős, Alfréd Rényi, and Vera T. Sós (1966), see Brendan's comment.

At first, if there exists a vertex of degree $k+1$, it is straightforward that $G$ is complete graph on $k+2$ vertices. If some vertex is joined with all other vertices, any other vertex has degree $k+1$, so we again get complete graph on $k+2$ vertices. Proceed with other cases.

At first, I consider a special case: $V=V_1\sqcup V_2$ for non-empty sets $V_1,V_2$, and all edges between $V_1$ and $V_2$ exist. Denote $n_1=|V_1|$, $n_2=|V_2|$, we have $n_1>1$, $n_2>1$, since there is no vertex joined with all other vertices. Choose vertices $v_i\in V_i$, $i=1,2$. Denote by $d_i$ their degrees in subgraphs $G(V_i)$. We have $d_1+d_2=k$, in particular, this does not depend on how we choose vertices. Next, any two vertices in $V_1$ have $k-n_2=d_1+d_2-n_2:=\alpha_1$ common neighbors in $V_1$. Let $W$ be a set of neighbors of $v_1$ in $G(V_1)$. Then counting edges between $W$ and $v_1\setminus (W\cup v_1)$ leads to a standard relation $(n_1-d_1-1)\alpha_1=d_1(d_1-\alpha_1-1)=d_1(n_2-d_2-1)>\alpha_1(n_2-d_2-1)$, i.e. $n_1-d_1-1>n_2-d_2-1$, analogously $n_2-d_2-1>n_1-d_1-1$. Contradiction.

Now assume that our graph is not regular and $1,\dots,m$, $n>m>1$, are all vertices of the same degree $d$ (we may always assume so, since any graph with at least 2 vertices has two vertices with the same degree.) Let $A$ be adjacency matrix, then consider eigenspace of $A^2$ corresponding to eigenvalue $d-k$. It is easy to see that this is a space of dimension $m-1$, it consists of vectors $(x_1,\dots,x_n)$ for which $x_{m+1}=\dots=x_n=0$ and $x_1+\dots+x_{m}=0$. But eigenspace of $A^2$ is invariant for $A$. Thus any vector with these properties is invariant for $A$. It follows that any vertex $i>m$ is either joined with all $1,\dots,m$, or with none of them. Assume that there are $s$ vertices joined with all $1,\dots,m$. Then $s\leqslant k$. since they are common neighbors of $1,2$. By already considereв case there is another vertex $v$, not joined with $1,\dots,m$. But 1 and $v$ have at most $s$ common neighbors. Thus $s=k$. But vertex 1 has degree at least $k+2$, so it is joined with some vertices $u_1,u_2\in\{1,\dots,m\}$. These two vertices have at least $k+1$ common neighbors. Contradiction.