An exercise in de Feo's isogeny notes asks for a proof that the $m$th division polynomial is divisible by $y$ if and only if $m$ is even, and none are divisible by $y^2$.
Given the recurrence relation $$\psi_{2n}=\frac{1}{2y}\psi_n(\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2),$$
then it's clear by induction (up to $2n-1$) that whether $n$ is even or odd, there are at least two factors of $y$ on the right-hand side, and one gets cancelled, so there is at least one left. However, I don't see any reason why $\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2$ doesn't "create" an extra factor of $y$ via subtraction.
What I mean is: If $\psi_{n+2}\psi_{n-1}^2=x^4$ and $\psi_{n-2}\psi_{n+1}^2=x^4-x^3-ax-b$, and the elliptic curve is $y^2=x^3+ax+b$, then $\psi_{n+2}\psi_{n-1}^2-\psi_{n-2}\psi_{n+1}^2=x^3+ax+b=y^2$. So for some reason these particular polynomials can't occur, but I don't see why.
I'm assuming "divisible" here means divisible in the ring $K[x,y]/\langle E(x,y)\rangle$ where $E(x,y)=y^2-x^3-ax-b$.