Let $G=(V,E)$ be a connected simple undirected graph and let $k>0$ be an integer such that

1. $\delta(G) \geq k$ (that is every vertex has at least $k$ neighbours), and
2. $K_{k+1}$ is not a minor of $G$.

Question: What is a lower bound for $|V|$ in terms of $k$? Of course I'm looking for a lower bound that is as high as possible.

(Remark: a trivial lower bound is $|V|\geq k+2$ as for $|V|=k+1$ we would get $G\cong K_{k+1}$ by point 1 above, violating point 2.)

Let $G=(V,E)$ be a connected simple undirected graph and let $k>0$ be an integer such that

1. $\delta(G) \geq k$ (that is every vertex has at least $k$ neighbours), and
2. $K_{k+1}$ is not a minor of $G$.

Question: In terms of $k$, how many vertices does a graph satisfying 1. and 2. above to contain at least? In other words, I am looking for an interesting lower bound for $|V|$.

(Remark: a trivial lower bound for $|V|$ is $|V|\geq k+2$ as for $|V|=k+1$ we would get $G\cong K_{k+1}$ by point 1 above, violating point 2.)