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We have $8a^2 + 1 = b^2$, so $b^2 - 8a^2 = 1$.

Consider the equation modulo $3$:

$8a^2 + 1 \equiv b^2 \pmod{3}$

Since $8 \equiv 2 \pmod{3}$, we have:

$2a^2 + 1 \equiv b^2 \pmod{3}$

Now, $a^2$ modulo $3$ can be $0$ or $1$:
- If $a \equiv 0 \pmod{3}$, then $a^2 \equiv 0$.
- If $a \equiv 1 \pmod{3}$ or $a \equiv 2 \pmod{3}$, then $a^2 \equiv 1$.

Case 1: $a \equiv 0 \pmod{3}$

Then $2a^2 + 1 \equiv 0 + 1 = 1 \pmod{3}$, so $b^2 \equiv 1 \pmod{3}$.

Possible values for $b$ modulo $3$ are $1$ or $2$ (since $1^2 \equiv 1$, $2^2 \equiv 1$).

Thus, $ab \equiv 0 \pmod{3}$.

Case 2: $a \equiv 1 \pmod{3}$ or $a \equiv 2 \pmod{3}$

Then $a^2 \equiv 1$, so $2a^2 + 1 \equiv 2 \cdot 1 + 1 = 3 \equiv 0 \pmod{3}$, so $b^2 \equiv 0 \pmod{3}$.

Thus, $b \equiv 0 \pmod{3}$, so $ab \equiv 0 \pmod{3}$.

In all cases, $ab$ is divisible by $3$.