Let $|l,m\rangle$ be standard angular momentum basis. I come across this identity
$$\langle2,-1|z|1,-1\rangle=\frac{\sqrt{3}}{2}\langle2,0|z|1,0\rangle$$ Using spherical harmonics, I can see this is indeed correct, but I am wondering can we deduce the coefficient $\sqrt{3}/2$ without explicitly calculating the spherical harmonics?
I considered using ladder operators to connect the two but didn't quite work out.
Yes. You can do this using the Wigner-Eckart theorem, which would give
\begin{align}
\langle 2,-1\vert z\vert 1,-1\rangle &=\frac{\langle 2\Vert r\Vert 1\rangle}{\sqrt{5}}C^{2,-1}_{1,0;1,-1}\, ,\\
\langle 2,0\vert z\vert 1,0\rangle &=\frac{\langle 2\Vert r\Vert 1\rangle}{\sqrt{5}}C^{2,0}_{1,0;1,0}
\end{align}
so that
\begin{align}
\frac{\langle 2,-1\vert z\vert 1,-1\rangle }{\langle 2,0\vert z\vert 1,0\rangle} = \frac{C^{2,-1}_{1,0;1,-1}}{C^{2,0}_{1,0;1,0}}=\frac{1/\sqrt{2}}{\sqrt{2/3}}=\frac{\sqrt{3}}{2}\, .
\end{align}
The reduced matrix element $\langle 2\Vert r\Vert 1\rangle$ and the dimensionality factor nicely cancel from the ratio and you're left with a ratio of Clebsch's.
The ladder operators (at least the angular momentum ones) cannot connect states with different $\ell$'s so you cannot use this to connect $\ell=2$ states and $\ell=1$ states. The "laddering" is done through the tensorial nature of the $\{\hat x\pm i \hat y,\hat z\}$ operators: as components of a $L=1$ tensor, their action on $\ell$ states can connect the initial states of angular momentum $\ell$ to states with $L'=\ell+1,\ell,\ell-1$ through $\ell\otimes 1=\ell+1\oplus\ell\oplus \ell-1$.
