In quantum algorithms I often find this identity: $$\frac{1}{2^n}\sum_{z\in\{0,1\}^n} (-1)^{z\cdot (x\oplus y)}=\delta_{xy}$$ where $x\oplus y$ is the bitwise sum.
I am not able to prove in general the case in which $x\neq y$, I can only show that it works for particular sets (e.g. $\{0,1\}^2$, $\{0,1\}^3$).
On
As mentioned in the comments, the key is to distribute and distribute and distribute: \begin{align} \frac{1}{2^n}\sum_{z\in\{0,1\}^n} (-1)^{z\cdot (x\oplus y)} & = \frac{1}{2^n}\sum_{z\in\{0,1\}^n} (-1)^{\sum_i z_i(x_i\oplus y_i)} \\ & = \frac{1}{2\stackrel{n}{\cdots}2} \sum_{z_1\in\{0,1\}} \cdots \sum_{z_n\in\{0,1\}} (-1)^{z_1(x_1\oplus y_1)} \cdots (-1)^{z_n(x_n\oplus y_n)} \\ & = \left( \frac12 \sum_{z_1\in\{0,1\}} (-1)^{z_1(x_1\oplus y_1)} \right) \cdots \left( \frac12 \sum_{z_n\in\{0,1\}} (-1)^{z_n(x_n\oplus y_n)} \right) \end{align} In other words, the problem reduces to a product of copies of the $n=1$ case, which you've solved already.
Suppose $x\ne y$, and let $a:=x\oplus y.$ Then $a$ has a leftmost $1$-bit, say in position $p$. For any $z$ whose $p$th bit is $1$ (i.e. matching $a$ at that bit), let $f_a(z)$ be the result of flipping that bit in $z$ (so as to not match $a$ at that bit). Now, among the $2^n$ $z$ values, exactly half of them match the leftmost $1$-bit of $a$ and half do not; furthermore, $z\cdot a$ and $f_a(z)\cdot a$ have opposite parity, hence $$\begin{align}\sum_{z\in\{0,1\}^n} (-1)^{z\cdot a}&=\sum_{z\in\{0,1\}^n:\text{$z$ matches the leftmost $1$-bit in $a$}} \left((-1)^{z\cdot a}+(-1)^{f_a(z)\cdot a}\right)=0. \end{align}$$