I disagree slightly with mr fantastic- equating the normal vector to the given vector will give a point that is NOT on the surface. Since you were asked to find points on the surface where the tangent plane is perpendicular to the vector <1, 0, 2>, set the normal vector equal to a multiple of that vector, say <a, 0, 2a>. That will give a "curve" of points, two of which are on the surface.
The attempt at a solution:
So, the gradient would be: <2x,2y,-2z>, but what do I do with the gradient vector to find these points?
The gradient vector at each point is perpendicular to the tangent plane at that point. Since you want the tangent plane to be perpendicular to <1, 0, 2>, you want a point (x, y, z) on such that <2x, 2y. -2z> is a multiple of that- such that < 2x, 2y, 2z>= <a, 0, 2a> for some number a.