You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin.

You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$\sum_{i=1}^{n+2} a_i v_i=0.$$ This gives enough room for manoeuvre to ensure some $a_i>0$ and some $a_j<0$. Thus we get some nontrivial relation $$\sum_{i\in I}a_i v_i=\sum_{j\in J}b_j v_j\qquad\qquad(*)$$ where all the $a_i>0$ and $b_j>0$ and $I$ and $J$ are disjoint non-empty sets of indices. It follows that the dot product of the two sides of $(*)$ is negative, but that contradicts it being the square of the length of the left side.

You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin.

You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$\sum_{i=1}^{n+2} a_i v_i=0.$$ This gives enough room for manoeuvre to ensure some $a_i>0$ and some $a_j<0$. Thus we get some nontrivial relation $$\sum_{i\in I}a_i v_i=\sum_{j\in J}b_j v_j\tag{$*$}\label{star}$$ where all the $a_i>0$ and $b_j>0$ and $I$ and $J$ are disjoint non-empty sets of indices. It follows that the dot product of the two sides of \eqref{star} is negative, but that contradicts it being the square of the length of the left side.