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.
The underlying issue here is that (assuming you want to stay within the real numbers) when $c<0$, the function $c^x$ is undefined for most values of $x$. Specifically, it's undefined unless $x$ is a rational number whose denominator is odd. There is no continuous/differentiable function underlying the places where it is defined.
Therefore, there is no possible guess-and-check algorithm that gradually becomes more accurate. First, guess-and-check algorithms require an underlying continuous function. Second, the value you're seeking might simply not be defined.
So the need to determine whether the exponent is a fraction with odd denominator, which in other contexts might be considered inelegant, here is simply a necessary step in the problem you're trying to solve. (And really, people shouldn't be inputting $c^x$ when $c<0$ and $x$ is a decimal ... they're just asking for trouble, for all the reasons mentioned above.)
Best Answer
You could also solve it like this.
Log both sides.
$\log 2^x = \log 10^{80}$
Using $\log a^r = r \cdot \log a$ (a rule of logarithms) the expression above becomes
$x \cdot \log 2 = 80 \cdot \log 10$
$\log 2$ & $\log 10$ are simply constants that you can plug into a calculator, so the expression becomes
$x = \frac{80\cdot \log 10}{\log2} \implies x = \frac{80}{\log 2}$
(since $\log 10 = 1$)