Define two prime powers $p^m<q^n$, ($m,n\in\mathbb N$) to be consecutive if it doesn't exist a prime power $r^k$ such that $p^m\!<r^k\!<q^n$. It seems that every even number is the difference between two consecutive prime powers, I haven't ran any large tests yet, but what about the odd numbers?
Are there two consecutive prime powers $p^m<q^n$ and $i\in\mathbb N^+$ such that $q^n-p^m=11^i$?
I have tested all consecutive prime powers less than $10,000,000$ without finding any solution to $q^n-p^m=11^i$.
(A pattern seems to suggest that no odd numbers $N\ge 31$ have a solution to $q^n-p^m=N$, consecutive prime powers. But this must be wrong since prime powers has similar asymptotic distribution as the primes).