According to Wolfram Alpha, $1^z=1$ for $z\in\mathbb{C}$. If this is true, then what is wrong with the following argument that $1^z$ has infinitely many values?
Let $z=x+iy$. Then, \begin{align} 1^z &= 1^{x+iy} \\ &= 1^x \cdot 1^{iy} \\ &= 1^{iy} \\ &= e^{iy\log(1)} \\ \log(1) &= \{0,2i \pi, 4i \pi,6i \pi, \ldots\} \\ 1^z &= \{e^{iy \cdot 0},e^{iy \cdot 2 i \pi},e^{iy \cdot 4 i \pi},e^{iy \cdot 6 i \pi},\ldots\} \\ &= \{e^0,e^{-2y\pi},e^{-4y\pi},e^{-6y\pi},\ldots\} \end{align}
Only one of these values is equal to $1$.
If, when $a>0$, you define $a^z$ as $\exp\bigl(z\log(a)\bigr)$, where $\log(a)$ is the only real logarithm of $a$, then indeed we always have $1^z=1$.
Otherwise, it's up to you to tell us how you are defining $1^z$. However, note that the set of all logarithms of $1$ is not $\{0,2\pi i,4\pi,i,6\pi i,\ldots\}$; it's $\{0,\pm2\pi i,\pm4\pi i,\pm6\pi i,\ldots\}$.