If $A^2-2A=0$, then $p(A)=0$, for $p(x)=x^2-2x$,
and thus the minimal polynomial of $A$ can by only one of the following three:
$$
x^2-2x,\,\,\, x\,\,\, \text{or}\,\,\, x-2,
$$
each of which possesses only simple roots, and thus $A$ is diagonalisable.

In general, if a matrix is annihilated by a polynomial which does not have any multiple roots, then this matrix is diagonalisable.

If you are not aware of the above fact, try Jordan decomposition, and see that $A$ can not have any Jordan blocks, in its Jordan decomposition.

**Note.** This proof works only in the case in which $0\ne 2$ in the field $\mathbb F$. If $2=0$ in $\mathbb F$, then the polynomial $p$ becomes $p(x)=x^2$, in such case the condition $A^2=0$, does not guarantee that $A$ is diagonalizable. In fact, there exist non-diagonalizable matrices satisfying $A^2=0$.