The following is Corollary 2.4.9 from the book "The Representation theory of the Symmetric Group" by James and Kerber.
For partition $\alpha,\beta \vdash n$, $\chi_{\alpha}(\beta)\neq 0 \implies \beta \leq (h_{11}^{\alpha}, h_{22}^{\alpha},\ldots)$. Here $\chi_{\alpha}$ is the irreducible character of $S_n$ corresponding to $\alpha$ and $\leq$ denotes the dominance order and $h_{ij}^{\alpha}$ is the hook length at $(i,j)$-th node of the Young diagram of $\alpha$.
The proof is not given in the book probably because it is not difficult, but I haven't been able to find a proof.
Here is what I tried : Let $\beta \not\leq (h_{11}^{\alpha}, h_{22}^{\beta},\ldots)$, then we want to prove that $\chi_{\alpha}(\beta)=0$. The idea is to show that $\beta$ contains a cycle whose length doesn't appear as a hook length of some node in the Young diagram of $\alpha$. Then the recursive Nakayama-Murnaghan rule immediately yields the result.
Also notice that for any partition $\alpha$, $h_{11}^{\alpha}$ cannot be too small. Let $P(n)$ be the set of all partition of $n$. Then we can consider the map $\Delta:P(n)\to P(n)$ given by $\alpha \mapsto (h_{11}^{\alpha}, h_{22}^{\alpha},\ldots)$. If $n=6$, one can check that the image of $\Delta$ is the set $\{(6), (5,1), (4,2)\}$. This shows that the least possible value of $h_{11}^{\alpha}$ is 4. In general, it seems that for any $\alpha \vdash n$, $h_{11}^{\alpha}\geq \frac{n}{2}+1$ if $n$ is even and $h_{11}^{\alpha}\geq \frac{n+3}{2}$ if $n$ is odd. From this it seemed to me that if $\beta=(\beta_1,\beta_2,\ldots)$ and $\beta \not\leq (h_{11}^{\alpha},h_{22}^{\alpha},\ldots)$, then $\beta_1>h_{11}^{\alpha}$. If it were to be true then since $h_{11}^{\alpha}$ is the highest hook lenght in $\alpha$, it would have followed that $\beta_1$ cannot be a hook length in the diagram of $\alpha$. But the conclusion that $\beta_1>h_{11}^{\alpha}$ can be easily seen to be false. For example - If $\alpha=(5,3,3)\vdash 11$, then taking $\beta=(6,5)$, it is clearly seen that $\beta\not\leq (7,3,1)=(h_{11}^{\alpha},h_{22}^{\alpha},h_{33}^{\alpha})$ but $\beta_1=6<h_{11}^{\alpha}$.
One can deduce some more restriction on the partition $(h_{11}^{\alpha}, h_{22}^{\alpha},\ldots,)$ . For example $h_{ii}^{\alpha}\leq h_{jj}^{\alpha}-2$ where $j=i-1$ and so on. But with all these I haven't been able to get the correct argument. I think there is some easy observation that I am overlooking. Any help will be highly appreciated. Thanks in advance.