Showing there is a node in the graph with one and only one edge

We have an undirected simple graph with $n$ vertices where for every pair of vertices $v_1,v_2$, if $d(v_1)=d(v_2)$ then the set of neighbours of $v_1$ is disjoint from the set of neighbours of $v_2$. Assuming the graph contains at least one edge, prove that there is a vertex of degree exactly $1$ in the graph.

For example the following graph has vertices of degree exactly $1$:

While this problem concerns a graph, I feel like there is a way to apply pigeonhole theory to prove this. Is this possible?

Solution 1:

Let $v$ be the vertex of the greatest degree and let $\operatorname{deg}(v)=k>0$. Let $N(v)$ be neighbors of vertex $v$. Then the degrees of all vertices from $N(v)$ are pairwise distinct, and if there are no vertices of degree 1 among them, then there must be a vertex of degree $k+1$ or more. This contradicts the choice of vertex $v$.