If $\{a,b,c\}$ is a basis for the Lie algebra $A_1$ with $[b,a]=2a, [b,c]=-2c$ and $[a,c]=b$. Suppose $V$ is a 13-dimensional $A_1$ module and that $\dim\{v\in V\mid bv=4v\}=\dim\{v\in V\mid bv=5v\}=1$. I want to show that $\dim\{v\in V\mid bv=8v\}=0$.
Taking $v\in \{v\in V\mid bv=8v\}$ then $b(cv)=[b,c]v+c(bv)=-2cv+c8v=6cv$. By the same considerations then $b(c(cv))=4(c(cv))$, so we may conclude $c(cv)\in \{v\in V\mid bv=4v\}$. Now I'm thinking, that if I somehow could conclude then $c(cv)$ is zero in $\{v\in V\mid bv=4v\}$, then $c(cv)=0$ giving that $v=0$, since $c$ is a basis element? However I'm kinda stuck right now, and can't get any further.
It seems pretty easy to solve, by using that $A_1$ can be classified with $sl(2)$ - then the dimension equals the corresponding nullspace. However I'm still curious to solve it without this classification.
This follows from the representation theory of $\mathfrak{sl}_2$, covered in any book on Lie algebras and representation theory; cf. https://math.stackexchange.com/a/3344350/96384 and https://math.stackexchange.com/a/3288482/96384. The element you call $b$ is often called $h$ in those sources.
Namely, since the weight spaces to the weights $4$ and $5$ are not zero, they generate submodules of dimension at least $5$ resp. $6$ respectively; if there were a non-zero weight vector to the weight $8$, then there would also be weight spaces to the weights $-8$ as well as $\pm 6$, i.e. at least four more dimensions. But $4+11 > 13$.