Infinite number of composite pairs $(6n + 1), (6n -1)$

Hint: Instead of even powers of $10$, try odd powers of $6$.

Try to make the expressions fit forms that you know factor. Sums and differences of cubes factor. So $n=6^2k^3$ will make the expressions take on those forms.

Here is a way of solving the problem "non-constructively"

You can find infinitely many disjoint sequences of $8$ consecutive composite numbers, and among any 8 consecutive numbers you can find a pair of the type $6n \pm 1$.

Once you figure this "non-constructive" solution, the standard examples lead to simple solutions. For example, it is easy to figure that $k!+5, k!+7$ work for all $k \geq 7$... This would lead to $n= \frac{k!}{6}+1$.