$a^2+ab+b^2=1\Leftrightarrow a^2+ab+b^2-1=0$. We must have that $b^2-4(b^2-1)=4-3b^2=k^2,k\in \mathbb{N}$.Since $k^2>0$ we have $b=0,\pm 1$ $b=0\Rightarrow a^2=1\Leftrightarrow a=\pm 1$ $b=1\Rightarrow a^2+a=0\Leftrightarrow a=0,-1$ $b=-1\Rightarrow a^2-a=1\Leftrightarrow a=0,1$. So $(a,b)=(1,0),(-1,0),(0,1),(-1,1),(0,-1),(1,-1)$.

Rewrite as $a^2 + ab + b^2-1 = 0$. Solve for $a$ and get a bound using a discriminant $a=\frac{-b\pm\sqrt{b^2-4\cdot (b^2-1)}}{2} \implies -3b^2+4\ge 0\implies b=\{-1,0,1\}.$ Casework. If $b=0$, then $a^2=1\implies a=1 \text{ or } a=-1$. If $b=-1$, then $a^2-a=0\implies a=0 \text{ or } a=1$. If $b=1$, then $a^2+a=0\implies a=0 \text{ or } a=-1$. Answer. $\boxed{(a,b)=(1,0);(0,1);(1,-1);(-1,1);(0,1);(0,-1)}$.