If you let $m=a^2$ then $n = (m+1)^2 - m = (a^2+1)^2 - a^2 = (a^2-a+1)(a^2+a+1).$

There are lots of choices for $a$ which make these last two factors simultaneously prime. E.g., $a= 2, 3, 6, 15, 21, \ldots.$ Both factors are in your range, if the range is a closed interval. For $a=6$, $m=36$ and $n=31\cdot 43.$ The factor $31$ is right on the boundary of your interval.