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][1] 


  [1]: http://www.wolframalpha.com/input/?i=solve+x+x+%2B+y+y+%2B+z+z+-+x+z+-+y+z+%3D%3D+1+over+the+integers