Two endomorphisms, $u$ and $v$, of $\mathbb{C^n}$ such that $[u,v] = \alpha u + \beta v$ are both triangular in the same basis

This question is on my course of Lie Groups. I am struggling on finding a better proof than the one I suggest below:

Let $V$ be a $\mathbb{C}$ vector space of dimension $n$. Let $u,v \in End(V,\mathbb{C})$ such that $[u,v] = \alpha u + \beta v$ where $\alpha , \beta \in \mathbb{C}$. Show that there exists a basis in which the matrices of u and v are triangular.

My Attempt:

There exists a basis B such that the matrix of $[u,v]$ is triangular. Then,

$[u,[u,v]] = u(\alpha u + \beta v) - (\alpha u + \beta v)u = \beta uv - \beta vu = \beta [u,v]$

$[v,[u,v]] = -\alpha [u,v]$

are also triangular. Let $w = [u,v]$, then all the elements below the diagonal in $uw$ are equal to those in $wu$. However, if $w$ is triangular then $uw$ must also be. Which implies that $wu$ is triangular (Idem for $v$). I am not sure on how to proceed from here. Does this imply that $u$ is triangular?

Do we have to go through each element: Let $(e_{n1})$ be the $n1$ entry of $wu$ which is equal to the dot product of the $n^{th}$ row of $w$ and the $1^{st}$ column of $u$. But, the $n^{th}$ row of $w$ can only have one entry which is non zero (i.e. $(w_{nn}$)), hence, ($u_{n1}$) must also be zero. We can apply the same argument for the whole $n^{th}$ row of $u$. Then we use recursion for the other rows to show that $u$ is triangular.

Is there something wrong with my proof? Is there another proof with is more elegant? I find my method a bit clumsy. Thank you for your help.

The vectors $u$ and $v$ span a $2$-dimensional solvable complex Lie algebra. By passing to the (faithful) adjoint representation we may assume that $u$ and $v$ are matrices of size $2$. Actually, this is assumed here already anyway. By Lie's theorem, there is a basis such that $u$ and $v$ are simultaneously upper-triangular.