Generally speaking, the problem arises because squaring is not a "reversible" operation. That is, while it is true that if $a=b$ then $a^2=b^2$, it is not true that if $a^2=b^2$ then $a=b$. (For instance, even though $(-1)^2=1^2$, it does not follow that $-1=1$)
This is in contrast to other kinds of equation manipulations that we use routinely when we solve equations. For example, if $a=b$, then $a+k=b+k$, and conversely: if $a+k=b+k$, then $a=b$. So we can add to both sides of an equation (for instance, you can go from $\sqrt{x+5}+1 = x$ to $\sqrt{x+5}=x-1$ by adding $-1$ to both sides) without changing the solution set of the equation. Likewise, we can multiply both sides of an equation by a nonzero number, because $a=b$ is true if and only if $ka=kb$ is true when $k\neq 0$. We can also take exponentials (since $a=b$ if and only if $e^a=e^b$) and so on.
But squaring doesn't work like that, because it cannot be "reversed". If you try to reverse the squaring, you run into a rather big problem; namely, that $\sqrt{x^2}=|x|$, and is not equal to $x$.
So when you go from $\sqrt{x+5} = x-1$ to $(\sqrt{x+5})^2 = (x-1)^2$, you are considering a new problem. Anything that was a solution to the old problem ($\sqrt{x+5}=x-1$) is still a solution to the new one, but there may be (and in fact are) things that are solutions to the new problem that do not solve the old problem.
Any such solutions (solutions to the new problem that are not solutions to the original problem) are sometimes called "extraneous solutions". Extraneous means "coming from the outside". In this case, it's a solution that comes from "outside" the original problem.
You can of course write
$$
(\sqrt{2x+1}+1)^2=x^2,
$$
but when you multiply out you get
$$
2x+2\sqrt{2x+1}+2=x^2,
$$
and there is still a radical in your new equation. The point in isolating the radical is that after that, as you square the equation, you get rid of it completely.
Best Answer
Usually when we write $\sqrt{x}$ where $x$ is a real number, we usually mean the principal root. This means the positive square root. While $(-4)^2=16$, we don't say that $\sqrt{16}=-4$, because $\sqrt{16}=4$, and if $\sqrt{16}$ were both values, it wouldn't be a function. This is why when squaring an equation can sometimes lead to extraneous solutions.
When taking the square root of an equation, we attach a $\pm$. So if $a^2=16$, we do $a=\pm\sqrt{16}=4,-4$.