If $N$ is normal in $G$, show $Z_{i}(G)N/N \leq Z_{i}(G/N)$ where $Z_{i}(G)$ is the $i$th term in the upper central series for $G$.
The argument is supposed to be by induction, and I have taken care of the base case.
Then suppose $Z_{i}(G)N/N \leq Z_{i}(G/N)$. By definition,
$$\begin{align} Z_{i+1}(G)N/N &= \{zN: z \in Z_{i+1}(G) \} \\ &= \{zN : [z,y] \in Z_{i}(G) \: \forall y \in G \} \\ &= \{ zN: (yz)^{-1}(zy)N \in Z_{i}(G)N/N\}. \end{align}$$
Then,
$$(yz)^{-1}(zy)N \in Z_{i}(G)N/N \implies (yz)^{-1}(zy)N \in Z_{i}(G/N) \leq Z_{i+1}(G/N) $$
by the inductive hypothesis, and then using the containment for upper central series. This is where I am stuck. How can I show $zN$ lies in $Z_{i+1}(G/N)$ from here?
As you note, if $K$ is a group, then showing that $x\in Z_{i+1}(K)$ is equivalent to showing that $[x,y]\in Z_i(K)$ for all $y\in K$.
You are assuming that $Z_i(G)N/N\leq Z_i(G/N)$. You want to prove that $Z_{i+1}(G)N/N\leq Z_{i+1}(G/N)$. That means that you want to show that for all $x\in Z_{i+1}(G)$, $[xN,yN]\in Z_i(G/N)$ for all $yN\in G/N$.
So, let $x\in Z_{i+1}G$. We want to show that $xN\in Z_{i+1}(G/N)$.
Let $y\in G$ be arbitrary. Then $[xN,yN] = [x,y]N$ in $G/N$. If $x\in Z_{i+1}(G)$, then $[x,y]\in Z_i(G)$, so $[xN,yN] = [x,y]N\in Z_i(G)N$. But $Z_i(G)N$. But $Z_i(G)N/N\leq Z_i(G/N)$ by the induction hypothesis, so $[xN,yN]\in Z_i(G/N)$.
As this holds for all $yN\in G/N$, it follows that $xN\in Z_{i+1}(G/N)$, as required.