Are these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ correct?
Might as well...
The power tower $x^{x^\ldots}$ is equivalent to the function $\exp(-W(-\log\,x))$, where $W(z)$ is the Lambert function, in the range $e^{-e}\leq x\leq e^{1/e}$ (as Norbert mentions in the comments; see also equation 13 in the MathWorld entry linked to). $\exp(-W(-\log\,x))$ can be inverted, like so:
$$\begin{align*} y&=\exp(-W(-\log\,x))\\ -\log\,y&=W(-\log\,x)\\ (\log\,y)\exp(-\log\,y)&=\log\,x\\ \frac{\log\,y}{y}&=\log\,x\\ x&=\exp\left(\frac{\log\,y}{y}\right)\\ x&=\exp\left(\log\,y^{1/y}\right)=y^{1/y} \end{align*}$$
If $y=2$, then $x=\sqrt2$.
Knoebel's paper establishes the interval of convergence $[\exp(-e),\exp(1/e)]$ for the power tower function, in the case of positive $z$. The paper notes that a full characterization of the region of convergence of $z^{z^\cdots}$ for complex $z$ remains to be done, but Thron, Shell (of Shellsort fame) and others have given partial results. See also this paper by Anderson for another discussion on the convergence of the power tower, this article by Cho and Park, where they discuss the inverses of the function $z^{1/z}$, and this article by Sondow and Marques.
If a solution exists, you have $$ x^2 = x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=2 $$ which means what you've got. (This part has been mentioned to be wrong for logarithmic properties misuse reasons) Not both of these are not solutions, since $$ 1 = \log_2(2) = \log_2(x^{x^{\dots}}) = x^{x^{\dots}} \log_2 (x) = 2 \log_2(x) $$ and in the case $x = -\sqrt 2$, $2\log_2(-\sqrt 2)$ is purely imaginary, thus cannot be $1$. (The logarithm of $\log_a(b^c) = c\log_a(b)$ part is the part that remains suspicious. As N.S. pointed out, I don't think this argument can be made right.)
One way to suggest $x=\sqrt 2$ would be to show that the sequence $$ x_n = \sqrt 2^{\dots^\sqrt2} $$ where exponentiation is taken $n$ times, is strictly increasing and bounded above by $2$. Numerical evidence suggests this : up to $n = 20$ I've seen that $x_n \le 2$ and $x_n$ is increasing. Convergence is slow and very long to compute though. I wasn't quite sure we could have convergence so I computed before finding a theoretical proof. Here's one : clearly $x_n$ is increasing, and $$ x_n^2 = \sqrt 2^{x_{n-1}} \times \sqrt 2^{x_{n-1}} = \sqrt 2^{2x_{n-1}} = 2^{x_{n-1}} \le 4 $$ by induction, so that $x^n \le 2$ for every $n$. Since the limit exists, it must be $\sqrt 2$.
Hope that helps,
(This should go as a comment but I doubt it would fit the box)
Also you should consider, whether you would better like to write $\small x$, $\small _bx $ , $\small _{_b}{_b}x $ ,$\small {_{...} } _{_b}{_b}x $ , because you always begin the evaluation at the top of the powertower and not at the bottom. And also then it is unambiguous to discuss $\small 2= 2 $, $\small 2 = _\sqrt22 $ , $\small 2= _{_\sqrt2}{_\sqrt2}2 $ and $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2 $ as evaluated from the top. It is then also correct to write $\small 4= 4 $, $\small 4 = _\sqrt24 $ , $\small 4= _{_\sqrt2}{_\sqrt2}4 $ and $\small 4= {_{...} } _{_\sqrt2}{_\sqrt2}4 $ as a second solution. (This is clearly no standard notation, but I really do not know why this did not become standard)
[added] Then one could also write $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}x \text{ for } -\infty \lt x \lt 4$ to note the convergence of all that initial values x, and because $\small x=\sqrt2 $ is in that range we can say $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}\sqrt2 $