Let $\pi$ be a representation a $\mathfrak{sl}(2)$. Let $v$ be a highest weight vector, with $\Lambda$ the highest weight. Then we must have $\pi(H)\pi(E^+)v=0$. But $\pi(H)$ has zero kernel, meaning that $\pi(E^+)v=0$. Since $\pi(E^+)$ also has zero kernel, $v=0$.
So this seems to show that there cannot be a highest weight vector. Of course this is wrong, and probably for a very silly reason, but I don't know what it is.