Answer: The two solutions are $(0, 0)$ and $(0, 2)$.

Since $2017^a$ is always odd, $b$ must be even, so $b = 2c$, $c$ integer. Therefore, $2017^a = 64(c^6 - c) + 1$ and thus $2017^a \equiv 1 \pmod{64}$. But we find $2017 \equiv 33 \pmod{64}$ and $2017^2 \equiv (1+32)^2 = 1+2\cdot32+32^2 \equiv 1 \pmod{64}$, so that the powers of $2017$ modulo $64$ alternate between $1$ and $33$. Therefore, $a$ is even and $2017^a$ is a perfect square. We denote the polynomial on the right-hand side of the given equation by $r(b) = b^6 - 32b + 1$ and show that it lies between two consecutive squares for $b > 4$:

Let $b > 4$. We have $r(b) < b^6 = (b^3)^2$ for $b > 0$. On the other hand, $r(b) > (b^3 - 1)^2$ because $b^6 - 32b + 1 > b^6 - 2b^3 + 1 \Leftrightarrow b > 4$. Since the square $2017^a$ is now between two consecutive squares, there are no solutions in this case.

Since $b$ is even, it remains to check $b = 4$, $b = 2$ and $b = 0$.

For $b = 4$, we regard the equation modulo $3$ and get $1 \equiv 1 - 2 + 1 = 0$, therefore, there is no solution in this case.

For $b = 2$, we get $2017^a = 2^6 - 2^6 + 1$, so we get the solution $(a, b) = (0, 2)$.

For $b = 0$, we get $2017^a = 1$, so we get the solution $(a, b) = (0, 0)$.

Therefore, $(0, 0)$ and $(0, 2)$ are the only solutions.