Let $M$ and $N$ be structures for a first order language $L$, with $M$ an elementary substructure of $N$. This means that $M$ is a substructure of $N$ and if $\varphi(x_1,\ldots,x_n)$ is a formula with free variables, and $m_1,\ldots,m_n$ are elements of $\operatorname{dom}(M)$, then $N \models \varphi(m_1,\ldots,m_n)$ iff $M \models \varphi(m_1,\ldots,m_n)$. Every source that I have looked at concludes that this implies that: if $N \models \exists x \varphi(x,m_1,\ldots,m_n)$, then $M \models \exists x\varphi(x,m_1,\ldots,m_n)$. (For example this implication is used in the proof of the Tarski-Vaught test.)
But doesn't $N\models \exists x \varphi(x,m_1,\ldots,m_n)$ mean only that there exists an $n$ in $\operatorname{dom}(N)$ (which a priori may not be in $\operatorname{dom}(M)$) such that $N \models \varphi (n,m_1,\ldots,m_n)$? Why does this imply that there is an $m$ in $\operatorname{dom}(M)$ such that $M\models \varphi(m,m_1,\ldots,m_n)$? What am I missing here?
If $M \prec N$, $m_1, \dotsc m_n\in M$ and $N\models \exists x\,\varphi(x, m_1, \dotsc m_n)$, then this same formula is true in $M$ of the elements $m_1, \dotsc m_n$. Because $M\models \exists x\,\varphi(x, m_1, \dotsc m_n)$, there is some $m\in M$ such that $M\models \varphi(m, m_1, \dotsc m_n)$. It follows that this is also true in $N$ of $m, m_1, \dotsc m_n$, namely, $N\models \varphi(m, m_1, \dotsc m_n)$.
It's true that $N$ may contain other witnesses to the existential statement that are not in $M$. Nevertheless, when $M$ is an elementary substructure of $N$ and $M$ models an existential statement with parameters in $M$, then $M$ must contain witnesses (like the element $m$, above) to that statement, which are also witnesses in $N$.