There is some number field $K$ containing $z$, and there we have a prime factorization $$ (z)=\frac{\mathfrak{p}_1\ldots\mathfrak{p}_a}{\mathfrak{q}_1\ldots\mathfrak{q_b}} $$ of fractional ideals, where no $\mathfrak{p}_i$ is equal to a $\mathfrak{q}_j$. If $nz^m$ is an algebraic integer, then $(\mathfrak{q}_1\ldots\mathfrak{q_b})^m$ contains $(n)$ (as ideals of $\mathcal{O}_K$). This means $$ N(\mathfrak{q}_1\ldots\mathfrak{q_b})^m|N(n)=n^{[K:\mathbb{Q}]}, $$ so that $$ n\geq \left(N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]}\right)^m. $$ This means we can take $$ c=N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]} $$

There is some number field $K$ containing $z$, and there we have a prime factorization $$ (z)=\frac{\mathfrak{p}_1\ldots\mathfrak{p}_a}{\mathfrak{q}_1\ldots\mathfrak{q_b}} $$ of fractional ideals, where no $\mathfrak{p}_i$ is equal to a $\mathfrak{q}_j$. If $nz^m$ is an algebraic integer, then $(\mathfrak{q}_1\ldots\mathfrak{q_b})^m$ divides $(n)$ (as ideals of $\mathcal{O}_K$). This means $$ N(\mathfrak{q}_1\ldots\mathfrak{q_b})^m|N(n)=n^{[K:\mathbb{Q}]}, $$ so that $$ n\geq \left(N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]}\right)^m. $$ We can take $$ c=N(\mathfrak{q}_1\ldots\mathfrak{q_b})^{1/[K:\mathbb{Q}]} $$