So let's now look at Larry's factorization (n^2+n+1) * (n^3-n+1) = 7^m.  Both polynomials are powers of 7 and for the values of n we are interested in we have n^3-n+1 > n^2+n+1.  But since both are powers of 7 then their quotient must also be a power of 7.

Then (n^3-n+1) / (n^2+n+1) = 7^j

Then dividing out the polynomial: n-1 - (n-2)/(n^2+n+1) = 7^j.  Then this implies (n-2)/(n^2+n+1) is an integer.  We are looking at n>=3, so the integer will be a positive integer. 

Then we can say n-2 > n^2+n+1.  This simplifies to -3 > n^2, which has no solutions; thus there are no solutions with n>=3.

The only positive integer solution of the equation is (n,m) = (2,2).