Is $\infty$ the solution to ${x}^{{x}^{{x}^{{x}^{x\dots}}}} = i$?
No! You've made a rather big error in your logic. Your algebraic manipulations prove an implication, not an equivalence - so $x^i=i$ has solutions that are not solutions to your original equation.
In particular, to hide any infinite weirdness, let's just write $f(x)=x^{x^{x^{\ldots}}}$ and not worry too much about how to define this*, except that we want $x^{f(x)}=f(x)$. You're trying to solve $$f(x)=i$$ So, the equation $f(x)=i$ then implies, by substituting $x^{f(x)}=f(x)$, that $$x^{f(x)}=i$$ which then, substituting $f(x)=i$ yields $$x^i=i.$$ This means that every solution to $f(x)=i$ is a solution to $x^i=i$. It does not mean that every solution to $x^i=i$ is a solution to $f(x)=i$. In particular, just because $e^{\pi/2}$ satisfies $x^i=i$ that doesn't mean it satisfies $f(x)=i$ - irrespective of whether we think $f(x)=\infty$ or not.
This is not a special problem with the question - one frequently encounters this issue. For instance, if I was trying to solve $$x+1=2$$ it is perfectly correct to square both sides and write $$x^2+2x+1=4$$ but one has to understand that the second equation has two solutions, which are $x=1$ and $x=-3$, and this doesn't imply that they solve the first equation. In general, algebraic manipulations are only valid in one direction, unless they can be undone by another manipulation (e.g. multiplying an equation by $2$ can be undone by dividing by $2$, but multiplying an equation by $x$ can't be undone if $x=0$ - which might lead to a new solution).
*Okay, if we were being a bit more careful, we might try to define $$f(x)=\lim_{n\rightarrow\infty}x^{x^{\ldots^x}}$$ where there are $n$ copies of $x$ in the tower. There some problems here - namely that this sequence might not converge, even if we had some meaningful** notion of $\infty$. We could prove that, wherever it does converge, we have the relation $$x^{f(x)}=f(x)$$ so a solution to $f(x)=i$ does truly imply $x^i=i$.
This sort of definition is pretty fragile though - there are cases that look okay, but act really badly. For instance, maybe if we were instead trying to solve $$x=1+\frac{1}x$$ we would try substituting and write $$x=1+\frac{1}x=1+\frac{1}{1+\frac{1}x}=1+\frac{1}{1+\frac{1}{1+\frac{1}x}}$$ which is a consequence of the the last equation (though, again, will create extraneous solutions if we just set the first and last equation equal). We'd be really tempted to write $$x=1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\ldots}}}$$ and to make the $x$ entirely vanish... but this doesn't work for subtle reasons - in particular, if we start with $1$ and make this substitution, the sequence $1,1+\frac{1}1,1+\frac{1}{1+\frac{1}1},\ldots$ actually does converge (to $\frac{1+\sqrt{5}}2$, which is a solution to $x=1+\frac{1}x$), but if we start with $\frac{1-\sqrt{5}}2$ - which is the other solution to $x=1+\frac{1}x$ - when we start substituting, we get a different answer - which tells us that our vanishing trick with the $x$ was not legal and that we would instead have to write out an honest limit. This issue arises pretty much universally when you have an "infinite" expression - you are basically doomed to say exactly what you mean, and then you might find out that the behavior of the whole object depends on what happened in the very beginning - which is, inconveniently, the part that you entirely conceal with "$\ldots$".
**While $\lim_{n\rightarrow\infty}x^{x^{\ldots^x}}$ goes to $\infty$ when $x$ is a real number greater than $1$, this is a fairly delicate statement that does not carry over well into the context of complex numbers mostly because definitions of such limits rely on the ordering of the real numbers. The complex numbers are not ordered, so the usual definitions simply don't apply. What's worse is that, in general, if a mathematician is talking about $\infty$ in the complex numbers, they far more commonly mean a concept that is completely different than what they meant with the symbol in the real numbers.