Where is the mistake in this argument? I cannot figure out where I am making a mistake:
suppose that $f$ is a function from $\mathbb{R}^n$, and $\nabla f(x) \cdot x=kf(x)$. Then we have
$$
kf(\alpha x)
=\nabla f(\alpha x) \cdot \alpha x
= \alpha \nabla f(x) \cdot \alpha x
= \alpha^2 \nabla f(x) \cdot x
= \alpha^2 kf(x).
$$
So given $k\ne 0$ we have $f(\alpha x)=\alpha^2 f(x)$.
Basically, I tried to prove $f(\alpha x)=\alpha^k f(x)$ given $\nabla f(x) \cdot x = k f(x)$. Am I actually making a mistake somewhere, or is it generally true that $f(\alpha x)= \alpha^2 f(x)$ given the condition? (so it is just my approach that is not working, then)

First note that $\nabla f(\alpha x) = \alpha \nabla f(x)$ will not hold in general. The gradients of $f$ at the points $x$ and $\alpha x$ do not need to be connected (do not mix this up with the gradients of $f$ and $x \mapsto f(\alpha x)$, but you do not consider the latter here; as you apply the given at the point $\alpha x$).
To your given problem: Define the function $g_x \colon (0,\infty) \to \mathbf R$ by $g_x(\alpha) = f(\alpha x)$. Then we have 
$$ g_x'(\alpha)\cdot \alpha = \nabla f(\alpha x) \cdot x \cdot \alpha = \nabla f(\alpha x)\cdot \alpha x = kf(\alpha x) = kg_x(\alpha) $$
Dividing by $g_x(\alpha)\alpha$, we get 
$$ \frac{g_x'(\alpha)}{g_x(\alpha)} = \frac k\alpha $$
or $$ (\log g_x)'(\alpha) = k \log' \alpha $$
giving 
$$ \log g_x(\alpha) = k\log \alpha + C $$
or $$ g_x(\alpha) = \exp(C) \cdot \alpha^k $$
