Given integer N, I wander is there is a method to find integers a,b,n,m > 1 such that $N = a^n - b^m$
When N is tiny and a,b,n,m are big the result sounds more interesting, like these examples
$2^{15} - 181^2 = 7$
$13^3 - 3^7 = 10$
$12^3 - 41^2 = 47$
$101^3 - 1015^2 = 76$
$6^7 - 529^2 = 95$
There is another relate question to the previous: can any integer N be written like difference of powers > 1 ?
$1 = 3^2 - 2^3$
$2 = 3^3 - 5^2$
$3 = 2^7 - 5^3$
$4 = 5^3 - 11^2$
$5 = 2^5 - 3^3$
$6 = ??$
$7 = 2^{15} - 181^2 = 2^7 - 11^2$
...