Generalisation for all numbers n, (n+1), (n+2) etc.

Consider small numbers of terms:

(n^4+n^2+1) = (n^2-n+1)(n^2+n+1)

(n^4+n^2+1)((n+1)^4+(n+1)^2+1) = (n^2-n+1)(n^2+n+1)^2(n^2+3n+3)

(n^4+n^2+1)((n+1)^4+(n+1)^2+1)((n+2)^4+(n+2)^2+1) = 

(n^2-n+1)(n^2+n+1)^2(n^2+3n+3)^2(n^2+5n+7)

(n^4+n^2+1)((n+1)^4+(n+1)^2+1)((n+2)^4+(n+2)^2+1)((n+3)^4+(n+3)^2+1) = 

(n^2-n+1)(n^2+n+1)^2(n^2+3n+3)^2(n^2+5n+7)^2(n^2+7n+13)

(n^4+n^2+1)((n+1)^4+(n+1)^2+1)((n+2)^4+(n+2)^2+1)((n+3)^4+(n+3)^2+1)((n+4)^4+(n+4)^2+1) =

(n^2-n+1)(n^2+n+1)^2(n^2+3n+3)^2(n^2+5n+7)^2(n^2+7n+13)^2(n^2+9n+21) 

The intermediate expressions, except for the first and last, form a series of squares, so the problem is equivalent to  (n^2-n+1)(n^2+(2k-1)n+(k^2-k+1)) = m^2 with {n,k} in the positive integers. 

But there are no such solutions:

{{k == -5, m == 7, n == 3}, {k == -3, m == 3, n == 2}, {k == -1, m == 1, n == 1}, {k == 0, m == 1, n == 0}, {k == 0, m == 1, n == 1}, {k == 0, m == 3, n == -1}, {k == 0, m == 3, n == 2}, {k == 0, m == 7, n == -2}, {k == 0, m == 7, n == 3}, {k == 1, m == 1, n == 0}, {k == 3, m == 3, n == -1}, {k == 5, m == 7, n == -2}}