Alternate solutions to algebraic equations? Most people have probably seen silly tricks in mathematics where people make the take simple steps and end up proving 1=0 or some similarly absurd result. As an example look at the (object/number/...?)
$e^{e^{e^{...}}}$
One might use some algebra to try to find the value of this doing the following
$e^{e^{e^{...}}}=S$
$\log(e^{e^{e^{...}}})=\log(S)$
$e^{e^{e^{...}}}\log(e)=\log(S)$
$e^{e^{e^{...}}}=\log(S)$
observing that the LHS is what we called S:
$S=\log(S)$ which has no real solution. But anyone can see that if we start with e and continually raise to a power of e it goes to infinity right? But we didn't find that solution with the algebraic approach presumably because some of the operations we did on infinity were not defined.
So how do we know there are no alternate solutions to simple equations like $x^2=4$ where the square root of the solution (in this case) is not defined?
Edit: to clarify, we all know to solve the equation above you can take the square root of both sides and end up with $\pm 2$. But what if there was an object that taking its square root had no meaning? Could (any) other properties still be well defined for this object?
 A: For your particular example, note that your two statements are not contradictory. To be explicit, it seems that your two claims are:

$S=\log(S)$ has no real solution, thus $S=e^{e^{e^{\ldots}}}$ cannot be assigned a sensible real value.

and

$S=e^{e^{e^{\ldots}}}$ tends to $\infty$ as more terms are added, thus $S=\infty$ is a reasonable value to adopt.

The fact that these are consistent with each other is readily seen: $\infty$ is not a real number, so our first statement says nothing about whether $S$ could be infinite. Indeed, if you extend the logarithm continuously to the extended reals, you get $\infty=\log(\infty)$. What this shows is that results derived symbolically are specific to the structure in which they are derived. So, if you are working within the reals, you must treat $x=\log(x)$ differently than if you are working in the extended reals.
This is the same case for the equation $x^2=4$. In the real numbers, you have the following property:

Every positive $y$ has exactly two $x$ such that $x^2=y$.

and you can use this to prove that $x=2$ and $x=-2$ are the only solutions.  More generally, that there are at most $2$ solutions is a property of integral domains, so these are still the only solutions in most reasonable extensions, like the complex numbers. However, if you change the domain to something more exotic, you can get more solutions. For instance, in the split-complex plane, you add a new "number" $j$ other than $1$ and $-1$ satisfying $j^2=1$. Then, $(2j)^2=4$ becomes a third solution to the given equation and $(-2j)^2=4$ becomes a fourth.
The point here is that, when you justify your algebraic manipulations, you are implicitly using things you know about the domain over which you are interested. The paradox you see is that, when you look at an equation over a larger domain, sometimes the assumptions you used in your algebra are no longer valid.
A: Note before answer: $x^2=4$ only has real solutions. $x=-2,2$.
I think what you mean are 'proofs' like this:
$x = y$. 
Then $x^2 = xy$. 
Subtract the same thing from both sides: 
$x^2 - y^2 = xy - y^2$. 
Dividing by $(x-y)$, obtain 
$x + y = y$. 
Since $x = y$, we see that 
$2 y = y$. 
Thus $2 = 1$, since we started with y nonzero. 
Subtracting $1$ from both sides, 
$1 = 0$.
This is nonsense since most (if not all) of these 'proofs' hinge on statements like $x=y$. Well then, when they divide by $x-y$, they are dividing by $0$ which is a BIG no-no in math.
Hope that helped!
A: I'm not entirely sure what you're asking.  Are you maybe saying this: 

Suppose $x$ is a solution to $x^2 = 4$.  Ordinarily we would solve this equation by taking square roots of both sides to get $x = \sqrt{4} = \pm 2$.  But what if it were the case that $x$ is a weird object for which taking the square root of $x^2$ does not make logical sense, much like the fact that it does not make sense to take the logarithm of $0$.  Would this mean that there could be other solutions to $x^2 = 4$?

If that's your question, I would answer by saying that there are ways to solve $x^2 = 4$ without taking the square root.  Personally, in unfamiliar situations, I try to avoid applying functions which aren't really functions in the strict sense anyway (the square root is multivalued; a function is supposed to give you exactly one output for each input).
Suppose that $x$ is a number such that $x^2 = 4$.  Since $a^2 - b^2 = (a-b)(a+b)$ for all numbers $a$ and $b$, we have that $$0 = x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)$$ If a product of two numbers is zero, then at least one of them has to be zero.  Thus either $x - 2$ is zero, or $x+2$ is zero.  Thus $x$ is either equal to $2$ or $-2$.
This shows that there are no other solutions to the equation $x^2 = 4$ other than $2$ or $-2$.  We have logically demonstrated that if $x$ is a number satisfying the equation $x^2 = 4$, then either $x$ or $2$ or $-2$.  So no clever trick could possibly demonstrate anything different unless it is fallacious.
Notice this does NOT show that $2$ or $-2$ actually solves the equation $x^2 = 4$.  All I showed was that if there is a solution, then it has to be $2$ or $-2$.
