Can Extraneous Roots be Introduced by Elimination? Suppose you have two equtaions:
$$2xy + y^2 = 0$$
$$x^2 + 2xy + 1 = 0$$
Subtracting the second from the first yields $y^2 - x^2 - 1 = 0$. Isolating y, we discover that $y = \pm\sqrt{x^2 + 1}$. However, by inspection we can wee that the $2xy$ term in both equations must be negative, which means that a single value of x cannot have both a positive and negative corresponding y-value. (i.e. if x is positive, then y must be negative). It seems that the equation $y = \pm\sqrt{x^2 + 1}$ contains an "extraneous root", but I'm struggling to wrap my mind around how that could be. After all, I solved for y w.r.t. x without squaring both sides. Could anyone help me understand what is going on?
The problem is that your reasoning isn't reversible. Your two equations together imply $y^2-x^2-1=0$, but the converse is not true: that equation does not imply your original system.

Compare your question to the following argument. The system $$x+y=1$$ $$2x+y=1$$
yields, by subtraction, that $x=0$. But this second equation allows $y$ to be arbitrary! So we get a whole lot of "extraneous solutions": $(0,y)$, for any $y$. The problem, of course, is that the equation $x=0$ does not, in turn, imply the original system. I take it you do not find this situation puzzling.

You generally lose information when you replace a system of equations with a linear combination of them. The combination will include the solutions of your original equation, but it likely will include non-solutions, as well. 

The form of the reasoning is: "any solution of the original equations will be a solution of this equation, too." Yes, but then you've only found a superset containing the solution set. To characterize the solution set exactly, you must worry about the converse.
You have taken a square root. Whenever you take a square root you can introduce an extraneous root. Even for the simple system $x^2=y^2$, $x=y$ if you took a square root of the first equation you'd get an extraneous root. You must always consider all the equations as they all contain information about the solution. If you had gone about solving it via elimination or substitution (rather than merging the two equations together randomly like you did) then you would have not encountered this.

If you solve your equation fully (and use complex numbers) you get two pairs of roots:
$$\left(\frac{1}{\sqrt3},-\frac{2}{\sqrt3}\right), \left(-\frac{1}{\sqrt3},\frac{2}{\sqrt3}\right),(i,0),(-i,0)$$
The second pair of roots does fit your reduced equation (with a small rearrangement), namely $x=\pm\sqrt{y^2-1}$.