What is wrong with this funny proof that 2 = 4 using infinite exponentiation?

First you need to determine what $x^{x^{x^{\cdots}}}$ means.

A natural attempt to give meaning to this would be the limit (if it exists) of the sequence $$ 1, ~x,~x^x,~x^{x^x}, ~x^{x^{x^x}}, \ldots $$ which can also be written as a recurrence

$$ z_0=1 \qquad z_{n+1} = x^{z_n} $$

If the limit exists at all, it must be a fixed point of the map $z\mapsto x^z$, but that doesn't mean that every fixed point has to be a limit.

In the case $x=\sqrt 2$, we easily see that both $z=2$ and $z=4$ are fixed points. The question is just which (if any) of them is the limit of the sequence. Plotting the function and using some iteration theory we see that $z=2$ is a stable fixed point and must be reached when we start from $1$, wheras $z=4$ is not stable -- if $z_n$ is close to $4$, the next $z_{n+1}$ will be less close to $4$.

You have already proved that IF there is any $x$ such that $x^{x^{\cdots}}=4$ when $x$ must be $\sqrt 2$. However, since $x=\sqrt 2$ doesn't in fact give $4$, then the conclusion is simply that $x^{x^{\cdots}}$ never takes the value 4.