I will show that $d$ is irrelevant, and for any such $d$, the set, $S_d=\{m: m\equiv 0\pmod{d}\}$ is the answer. Note that, if ${\rm gcd}(m,n)=d$, then $m\in S_d$ clearly. Now, let $m=dk$ with $k\in\mathbb{Z}^+$ arbitrary. Let $n=di$. We will establish that for any $k\in\mathbb{Z}^+$ there exists an $i$ such that: 1) $(i,k)=1$, and 2) $(4di+1,k)=1$. For $k=1$ this is trivial, thus suppose $k>1$; and let $p_1,\dots,p_\ell$ be the (distinct) prime divisors of $k$. Now, if $p_1=2$, choosing $i\equiv 1\pmod{2}$ ensures the first condition, and the second is trivially true. Suppose $p_j>2$ now. If there is a $j$ such that $p_j\mid d$, then $4di+1$ is coprime with $p_j$ for any choice. If $d$ is coprime with $p_j$, then simply choose $i$ so that $4di+1 \equiv 2\pmod{p_j}$. This is clearly doable as $p_j>2$. In any event, we have established the existence of $\alpha_1,\dots,\alpha_\ell$ so that $\alpha_j\not\equiv 0\pmod{p_j}$ and if $i\equiv \alpha_j\pmod{p_j}$, then $(4di+1,p_j)=1$, as well. Such an $i$ trivially exists due to Chinese remainder theorem. Thus, for any arbitrary $d$, $S_d$ is indeed the set of all such $m$'s enjoying the condition.