Why do I get incorrect solutions x=0 and x=2 for x+1+4x+1=0?

Step 1
Here is one correct way to solve this problem I learnt from comments.
I should write
${\displaystyle x+1=-\sqrt{4x+1}}$
$⟺\{\begin{array}{l}(x+1{)}^2=4x+1\\ (x+1)\le 0\end{array}$
$⟺\{\begin{array}{l}x(x-2)=0\\ x\le -1\end{array}$
Then I should solve ${\displaystyle x(x-2)=0}$ to get ${\displaystyle x=0}$ and ${\displaystyle x=2}$. Then I should reject both solutions because both solutions do not satisfy ${\displaystyle x\le 1}$. Therefore the equation has no solutions.

Step 1
You can easily track where the extra solutions were introduced
${\displaystyle x+1+\sqrt{4x+1}=0\to 0+1+\sqrt{1}=0}$
${\displaystyle x+1=-\sqrt{4x+1}\to 0+1=-\sqrt{0+1}}$
${\displaystyle x^2+2x+1=4x+1\to 0+0+1=0+1}$
${\displaystyle x^2-2x=0\to 0-0=0}$
${\displaystyle x(x-2)=0\to 0(0-2)=0}$
Repeat with ${\displaystyle x=2}$.
If you turn ${\displaystyle a=b}$ to ${\displaystyle a^2=b^2}$, you also allow ${\displaystyle a=-b}$.