Where's the problem with a false "proof": $\;1^0 = 1^2 \overset{?}\implies 0 = 2$ What's wrong with this:

$$\large 1^0=1^2$$

Since bases are same, therefore

$$\large 0=2$$

My thinking:

Since the function $\,f(x)=1^x\,$ is not one to one, therefore whenever $\,f(x)=f(y),\,$ $x\,$ need not be equal to $\,y$.

Question:

Is my reasoning sound?
Yes. That's fine reasoning. 

Indeed $f(x) = 1^x = 1 \;\forall x \in \mathbb R$. 

Certainly, as you note, $f$ fails to be injective, so it is NOT the case that $\forall x, y \in \mathbb R, \; f(x) = f(y) \implies x = y$.
The best possible explanation could be that  f(x)=1^x is not bijective hence making the inverse non existent and so x = y does not mean any thing.