I have to show for an even $q$ and $u \neq 0$ that the equation $X^2 + ux + v = 0, u,v \in \mathbb{F}_q$ is solvable over $\mathbb{F}_q$ iff $v/u^2$ is of the form $z^2+z$ for a $z \in \mathbb{F}_q$.
My progress so far:
"$\Leftarrow$"
Let $X =: bY \Rightarrow u^2Y^2+u^2Y+v=0 \Leftrightarrow u^2(Y^2+Y+v/u^2)=0 \Rightarrow X$ is the root of the equation $Y^2+Y+v/u^2 = 0$.
Therefore: $Y^2+Y+v/u^2 = 0 \Leftrightarrow Y^2+Y+z^2+z = 0$.
How can I continue or am I completely on the wrong path?
Your argument is correct. One can simply argue like the following:
Suppose that $z^2+z=v/u^2$. This is equivalent to $$u^2z^2 + u^2 z + v=0$$ which is another way to write $$(uz)^2+u(uz)+v=0$$ Hence, calling $x=uz$, you have $$x^2+ux+v=0$$
Hence, if the last equation has a solution $x$, then you have that $z^2+z=v/u^2$, where $z=u^{-1}x$.
On the other hand, if $z^2+z=v/u^2$, the equation $x^2+ux+v=0$ has a solution $x=uz$.