As percusse and GEdgar point-out in there comments that the reason this seemingly simple equation is not solvable using simple algebra lies in the fact that that the LHS of $2^x = x^2$ is a transcendental function. i.e. it cannot be expressed as a polynomial. Actually the closest it can come to in a "polynomial" form is its Maclaurin series form (see below).
Using pre-calculus techniques you can show, for instance, that you can take log of both sides as in
$$2^x = x^2$$
$$\implies ln(2^x) = ln(x^2) \quad \forall x \ne 0 $$
$$\implies x ln(2) = 2 ln(x) $$
$$\implies ln(x) = {2x \over ln(2)} \quad \textbf {(A)}$$
So the solution to our problem are all values of $x$ that are the roots of equation $\textbf{(A)}$ ... Pre-calculus you can use graphing techniques to determine the answer.
Solving transcendental functions, in general, requires a lot of different calculus techniques, that are probably beyond the scope of this answer.
Infinite Series for ${2^x}$
Using Taylor's Theorem (which is part of calculus) we can show that:
$$e^u = \sum_{n=0}^{ \infty } {u^n \over n!} = 1 + {u^1 \over 1!} + {u^2 \over 2!} + {u^3 \over 3!} + {u^4 \over 4!} + \cdots \quad \textbf{(B)}$$
For considerable historical reasons $\textbf{(B)}$ is called Maclaurin series for $e^u$. You can find Maclaurin series for a large number of functions that have certain properties.
For purpose of this discussion, assume that (B) is provable. We can use it to express the infinite series for $2^x$ by noting that $2 \equiv e^{ln(2)}$, and that $(a^x)^y = (a)^{xy}$.
$$ [2]^x = [e^{ln(2)}]^x = [e^{ln(2) \dot x}]$$
Substituting $u$ with $2^x$ in $\textbf{(B)}$ power-series we get:
$$2^x = ln(2) \sum_{n=0}^{ \infty } {x^n \over n!} = ln(2) \left( 1 + {x^1 \over 1!} + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} + \cdots \right)$$
As you can see solving (A) without knowing some more properties about behavior of $e^x$ becomes intractable. That is what calculus is all about ;) once you get into it, you wil see that these problems become solvable. Although, the solutions are, by no means, trivial.
This solution is not "elementary" as we say. You cannot get it using the operations from high-school algebra.
We can get the solution using the Lambert W function LINK
Definition: $u e^u = t$ iff $u = W(t)$. So reason like this
$$
\ln(x^2)=x
\\
x^2 = e^x
\\
x=\pm e^{x/2}
\\
xe^{-x/2} =\pm 1
\\
\frac{-x}{2}e^{-x/2} = \mp\frac{1}{2}
\\
\frac{-x}{2} = W(\mp 1/2)
\\
x = -2W(\mp 1/2)
$$
It turns out that $W(-1/2)$ is not real, but $W(1/2)$ is.
Then we get $x =-2W(1/2) \approx -0.703467$.
Best Answer
Considering $$y^2 = \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}$$ we could start defining $$z=1 - xe^{xy^2}\implies y^2=\frac{\log(z)}z$$ Solving for $z$ $$z=-\frac{W\left(-y^2\right)}{y^2}$$ where $W(.)$ is Lambert function. So $$1 - xe^{xy^2}=-\frac{W\left(-y^2\right)}{y^2}$$ and, again, Lambert function $$x=\frac{W\left(y^2+W\left(-y^2\right)\right)}{y^2}$$
Do not ask me for $y(x)$, please !
Edit
Since you mention the product logarithm, I suppose that you can run Mathematica. If so,ask for the contour plot of $$f(x,y)=y^2 - \frac{\ln(1 - xe^{xy^2})}{1 - xe^{xy^2}}=0$$ for $-\frac{1}{\sqrt{e}} \leq y \leq \frac{1}{\sqrt{e}}$ and $-1000 \leq x \leq 0$.