2 formatting (and trying to fix the link)

To elaborate on Robin's suggestion, set $x=v+w,y=u+w,z=u+v$, and the equation becomes $$u^2+v^2+2w^2=1,$$ with $u=(y+z-x)/2,v=(x-y+z)/2,w=(x+y-z)/2$ being half-integers. Now a brute force run through the possibilities is feasible, since $|u|\leq 1$ and such.

For the quick answer, you can use Mathematica:

Reduce[x x + y y + z z - x z - y z == 1, Integers]

or even wolfram|alpha:

1

To elaborate on Robin's suggestion, set $x=v+w,y=u+w,z=u+v$, and the equation becomes $$u^2+v^2+2w^2=1,$$ with $u=(y+z-x)/2,v=(x-y+z)/2,w=(x+y-z)/2$ being half-integers. Now a brute force run through the possibilities is feasible, since $|u|\leq 1$ and such.

For the quick answer, you can use Mathematica: Reduce[x x + y y + z z - x z - y z == 1, Integers] or even wolfram|alpha: solve x x + y y + z z - x z - y z == 1 over the integers