The question is to evaluate $$\sqrt2^{\sqrt2^{\sqrt2^{\:^{.^{\:^{.^{\: \infty}}}}}}}$$
Let $$x^{x^{x^{x^{\;^{\cdots{^\infty}}}}}}=2$$ Then $x^2=2 \implies x=\sqrt2$
Now Let $$x^{x^{x^{\:^{.^{\:^{{. ^\infty}}}}}}}=4$$ Then $x^4=4\implies x=\sqrt2$. So I am getting the two values $2$ and $4$ for $\sqrt2^{\sqrt2^{\sqrt2^{\:^{.^{\:^{.^{\: \infty}}}}}}}$
where is the ambiguity ?
In each case, you are tacitly assuming that a (positive) real number $x$ exists such that the sequence $x_0=1$, $x_{n+1}=x^{x_n}$ for $n=0,1,2,\ldots$, has a given limit (namely $2$ in the first case and $4$ in the second). That tacit assumption may or may not be valid in either case. What you've shown is that it cannot be valid in both cases.
As for the limit (if any...) of the sequence with $x=\sqrt2$, note first that $x_0=1\lt\sqrt2=x_1$, from which we find that $x_{n-1}\lt x_n$ implies $x_n=\sqrt2^{x_{n-1}}\lt\sqrt2^{x_n}=x_{n+1}$ for all $n$, so that the sequence is monotonically increasing (and strictly positive). Note second that $x_0=1\lt2$, from which it follows that $x_{n-1}\lt2$ implies $x_n=\sqrt2^{x_{n-1}}\lt\sqrt2^2=2$ for all $n$, so the sequence is bounded above (by $2$). Together these imply the sequence with $x=\sqrt2$ has a positive limit, and that limit is less than or equal to $2$.
If we now let $L$ denote that limit, then we can conclude that $L=\sqrt2^L$, or $L^2=2^L$ (with $L\ge0$). By graphing the parabola $y=L^2$ and the exponential curve $y=2^L$, it's easy to see that the only non-negative solutions to $L^2=2^L$ are $L=2$ and $L=4$. But we already know that $L\le2$. Hence the limit of the sequence with $x=\sqrt2$ is $2$.
