 Solution

Both $n$ and $m$ divide $b^{2p}-1$, hence so does $N=nm$ after checking $\gcd (n,m)=1$ under the hypothesis. Also $n\equiv p(\mathrm{mod}b-1)$ and the elementary factorization gives $N\equiv 1(\mathrm{mod}2p)$; part (a) with $k=2p$ applies. As odd primes $p$ grow, $N$ grows, yielding infinitely many examples.