The use of modulo in the mathematical induction proof of $7^{2n}-9$ being divisible by 2

Following your approach, note that in the inductive step, $$7^{2(k+1)}-9=49\cdot7^{2k}-9=49\underbrace{(7^{2k}-9)}_{\equiv 0 \pmod{2}}+\underbrace{(49-1)}_{\equiv 0 \pmod{2}}\cdot9\equiv 0 \pmod{2}.$$

P.S. Instead of induction, a straightforward proof follows from the basic facts that:

1) the product of two odd numbers is odd (so $7^{2n}$ is odd);

2) the difference of two odd numbers is even (so $7^{2n}-9$ is even).

As an alternative simply observe that

$$7^{2n}-9\equiv (1)^{2n}-1\equiv 0 \mod 2$$