Let $N=k^2$. Since $a+b+c=0$ we have $$N=2017+(a^2+ab+b^2)^2=k^2 \iff (k - a^2 - ab -b^2 )(k + a^2 + ab + b^2 ) = 2017$$Since $2017$ is prime $k - a^2 - ab - b^2 =1$ and $k + a^2 + ab + b^2=2017$ from which we get $k=1009$ and $ a^2 + ab + b^2=1008$. We see that $$a^2 + ab + b^2\equiv 0 \mod 18 \implies a=6n, b=6m$$So we have $$m^2+mn+n^2=28$$We see again that $$m^2+mn+n^2\equiv 0 \mod 2 \implies m=2x, n=2y$$Hence, $$x^2+xy+y^2=7$$Now the discriminant of this quadratic must be a perfect square in order integer solutions to exist. $$\Delta= 28-3x^2>0 \implies x^2=1,4,9$$Since $a>0$, $x>0$ as well. Therefore $x=1$ or $x=2$ or $x=3$. Because $y<0$ the possible solutions are $$(a,b)=(12,-36),(24,-36),(36,-24),(36,-12)$$Since $a>0>b>c$ and $a+b+c=0$ we get the only triplet $(a,b,c)=(36,-12,-24)$.

Let $d=-b$ and $e=-c$, so we now have $a=d+e$(*), $N=2017+a^3d-d^3e+e^3a=x^2$ for some integer $x$. Using (*) and doing some algebra, we get $x^2=d^4+e^4+3d^2e^2+2d^3e+2e^3d+2017=(d^2+e^2)^2+2ed(d^2+e^2)+(ed)^2+2017=(d^2+e^2+ed)^2+2017$. Let $s=d^2+e^2+ed>0$. We have $2017=(x+s)(x-s)$, and since $2017$ is a prime, the only pairs $(x,s)$ $(s>0)$ which satisfy the equation are ($\pm 1009,1008)$, which means $s=1008$. Now we have to solve $e^2+d^2+ed=1008$. Let $d=2^md_0$ and $e=2^ne_0$, with $e_0$ and $d_0$ being odd. Right side of the equation ($1008$) is equal to $2^4*63$, so $min(2m,2n)=4$ because if both were bigger than two the left side would be divisible by at least $2^6$, and if one of them was smaller than two then the left side wouldn't be divisible by $2^4$. WLOG we can assume $m=2$. If we divide both sides by $16$ we get $d_0^2+t^2+td_0=63$, with $t=\frac{e}{4}$. We know $d_0$ is odd, so we do casework for $d_0=1$, $3$, $5$ and $7$. Only for $d_0=3$ we get integer solutions-> we get $t=6$, $e=24$, $d=12$, $b=-12$, $c=-24$, $a=36$.