This solution was created joint with Skye Binegar.
Assume that these conditions hold of a graph $G$. I'm going to enumerate your properties as
- No vertex is adjacent to every other vertex.
- Any two non-adjacent vertices are both adjacent to a single unique vertex.
- No triangles exist.
Let $v$ be a vertex of highest degree, say $k$, and let $v_1,\dots, v_k$ be its neighbors. Note that by $(3)$, no two neighbors of $v$ are adjacent. By $(1)$, there is some vertex $x$ that $v$ is not adjacent to. Since $v$ and $x$ are not adjacent, by $(2)$ there is a unique $v_i$ such that $v_i$ is adjacent to $x$. Let's assume that $v_1$ is this vertex.
Now, for every $2\le i\le k$ we see that $v_i$ cannot be adjacent to $x$ by uniqueness of $v_1$ guaranteed by $(2)$. Therefore, for each $i$ there must exist some unique vertex $w_i$ adjacent to both $x$ and $v_i$ by $(2)$, since $v_i$ and $x$ are not adjacent when $i\ne 1$. Note that each $w_i$ must be distinct, as otherwise $v$ would be connected to some $w_i$ in two different ways, contradicting $(2)$.
This means that $x$ is adjacent to $v_1$ and $w_2,\dots,w_k$. Since $v$ has the highest degree $k$, and $x$ has degree at least $k$, this shows that $x$ has degree $k$. Therefore, any vertex that is not adjacent to a vertex of degree $k$ must also have degree $k$. Now, note that each $v_i$ is not adjacent to $x$, which we showed has degree $k$. Therefore, this argument shows that each $v_i$ must also have degree $k$.
Therefore, if a vertex has degree $k$ then so do all of its neighbors. By the connectivity of $G$ and maximality of $k$, this proves that $G$ is $k$-regular.