All $c\le2$ work: Let $k\ge1$ be an integer, and choose $n=k^2+1$ and $m=k^2-2k+2$. (1). Then $2n-m=(k+1)^2-1$, so that there is no squares among $n,n+1,\ldots,2n-m$. (2). Furthermore $m+2\sqrt{m-1}+1=(k^2-2k+2)+2(k-1)+1=k^2+1=n$. And $c>2$ does not work: (1). Let $k^2$ be the largest square below $n$; then $k^2+1\le n$ and $k\le\sqrt{n-1}$. (2). As there are no squares among $n,n+1,\ldots,2n-m$, we get $2n-m\le k^2+2k$. (3). This implies $2n-m \le (\sqrt{n-1})^2+2\sqrt{n-1}$, which can be rewritten into $\sqrt{n-1}\le\sqrt{m-1}+1$. (4). Squaring of $\sqrt{n-1}\le\sqrt{m-1}+1$ finally yields $n\le m+2\sqrt{m-1}+1$.