$(a, b) = (0, 0)$ and $(a, b) = (1, 0)$.
By adding $1$ to both sides of the given equation, we can rewrite it as
$$a^8 - a^7 + 1 = (19b + 1)^2.$$
Left hand side of this equation has the following factorization:
$$a^8 - a^7 + 1 = (a^8 - a^7 + a^6) - (a^6 - 1) = (a^2 - a + 1) \cdot (a^6 - (a+1)(a^3-1)).$$
Let $d = \text{gcd}(a^2 - a + 1, a^6 - (a+1)(a^3-1))$. Since $d|a^2 - a + 1$ we get $d|(a+1)(a^2 - a + 1)$ and $d|a^3 + 1$. Thus, $a^3 \equiv -1 \pmod d$. Therefore,
$$a^6 - (a+1)(a^3-1) \equiv 1+2(a+1) \equiv 2a+3 \pmod d.$$
Now since $d|a(2a+3)$ and $d|2(a^2-a+1)$ we get that $d|(5a-2)$. Finally, $d|2(5a-2)$ and $d|5(2a+3)$ yield $d|19$, which means that either $d=1$ or $d=19$. Note that since $d^2|(19b+1)^2$, $d=19$ is not possible.
Thus, $a^2 - a + 1$ and $a^6 - (a+1)(a^3 - 1)$ are coprime integers whose product is a perfect square. Therefore, each of them should be a perfect square. Note that $(a-1)^2 < a^2 - a + 1 < a^2$ for $a > 1$, and $(a-1)^2 > a^2 - a + 1 > a^2$ for $a < 0$. Hence, $a^2 - a + 1$ is not a perfect square except $a=0, 1$. Inserting $a=0$ and $a=1$ to the main equation we get two solutions: $(a, b) = (0, 0)$ and $(a, b) = (1, 0)$.