Let us consider the quadratic equation:
$$(m - 1)x^2 - 2mx + 2 = 0.$$
This equation has no real solutions if and only if its discriminant is negative.

The discriminant $D$ is:
$$D = [-2m]^2 - 4(m-1) \cdot 2 = 4m^2 - 8(m-1) = 4m^2 - 8m + 8.$$
We require:
$$4m^2 - 8m + 8 < 0.$$
Divide both sides by $4$:
$$m^2 - 2m + 2 < 0.$$
But the quadratic $m^2 - 2m + 2$ has discriminant:
$$(-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 < 0.$$
So $m^2 - 2m + 2$ is always positive for all real $m$.

Therefore, there is **no real value of $m$** for which the equation has no real solutions.