Before we end 2017......

Consider the equation over the ring $R = {\mathbb Z}[\omega],$ where $\omega$ is a primitive third root of unity. It is an important fact that $R$ is a unique factorization domain; any element can be factored as a product of irreducibles, unique up to ordering and scaling by a unit. There is a norm function $N : R \to \mathbb Z$ defined by $N(a-b\omega) = (a-b\omega)(a-b\omega^2) = a^2+ab+b^2,$ and the nice thing about it is that it is multiplicative: $N(\alpha\beta) = N(\alpha)N(\beta).$

Then we get $N(a-b\omega) = (a-b\omega)(a-b\omega^2) = 2017,$ and the prime $2017$ splits as the product of two irreducibles whose norm is $2017$ . (Anything whose norm is $2017$ must be irreducible because if it factored, the norm of one of the factors would have to be 1, so it would be a unit.)

By inspection, we can see $N(41-7\omega) = 41^2 + 7 \cdot 41 + 7^2 = 2017,$ so there is at least one solution. (For primes $p$ not divisible by 3, a solution exists if and only if $p \equiv 1$ mod $3,$ by some basic algebraic number theory machinery.)

How many other solutions are there? Well, the factorization $(41-7\omega)(41-7\omega^2) = 2017$ is unique up to switching the factors and multiplying by a unit. There are exactly six units in ${\mathbb Z}[\omega],$ consisting of the sixth roots of unity generated by $1+\omega$ (this can be verified by fancy arguments involving Dirichlet's unit theorem, or pretty easily by direct computation). So there are a total of $2 \cdot 6 = \fbox{12}$ distinct solutions. For completeness, I'll list them here:

$(\pm 41, \pm 7) \\ (\pm 48, \mp 41) \\ (\pm 48, \mp 7) \\ (\pm 7, \pm 41) \\ (\pm 41, \mp 48) \\ (\pm 7, \mp 48)$