Yes. By the "discrete intermediate value trick" (the number of such numbers changes only by $\pm 1$ when $m$ increases), it suffices to find infinitely many $m$ with at most $2017$ and infinitely many $m$ with at least $2017$. The first is easy: There are at most $n^{5/6}$ successful numbers up to $n$, so they have density $0$, in particular there is $m$ such that none of $m+1,\dots,m+2016^2$ are successful. On the other hand, choosing $m=k^6$, we see that all the numbers $m+1^2,m+2^2,\dots,m+2016^2$ are successful as $m$ is a third power. But also $m+8=(k^3)^2+2^3$ is successful (and indeed many more), so we are done.