$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$
I'm transcribing below (but see edit history for a scan) a calculation from pg 17 of this article on Lie groups and Lie algebras
$$ [N,X]=cX. \tag{6.1.1}$$ If $N$ has an eigenvector $\ket{n}$ which satisfies $N \ket{n} = n \ket{n}$, then we can apply $NX$ to $\ket{n}$ and use (6.1.1) to obtain \begin{align} NX \ket{n} &= \{XN + [N, X]\}\ket{n} \\ &= XN\ket{n} + [N, X] \ket{n} \\ &= Xn \ket{n} + cX \ket{n} \\ &= (n + c) X \ket{n} \end{align} Hence, \begin{align} N \ket{n} &= n \ket{n}\tag{6.1.2}\\ X \ket{n} &\propto \ket{n + c}\tag{6.1.3} \\ N \ket{n + c} &= (n + c)\ket{n + c}\tag{6.1.4} \end{align} So the action of $X$ on $\ket{n}$ is to create a new eigenvector of $N$, $\ket{n + c}$ which eigenvalue $n+c$. If $c$ is positive, then $X$ is a raising operator and if $c$ is negative it is a lowering operator.
What does $X|n\rangle \propto |n+c\rangle$ mean? They both seem to be vectors. Are they proportional as vectors? What about the previous calculation suggests this?