The fundamental problem is that when novice students (and even some non-novices ones) manipulate equations, they get the false impression that anything implied by their original expression will also imply their original expression.
To take an intentionally absurd example (absurd because we actually have the solution written down to start with), they reason like this:
$$x = 1$$
"square both sides"
$$x^2 = 1^2 = 1$$
The solutions of this are $1$ and $-1$ (true) and therefore (invalid deduction) the solutions of the original equation are $1$ and $-1$ (false conclusion).
If they were more careful they would write:
$x = 1 \implies x^2 = 1 \implies (x = 1) \vee (x = -1)$
So we know that nothing other than $1$ and $-1$ can be solutions, but we still don't know which (if any) is really a solution.
It's entirely analogous to reasoning:
$$x < 1$$
therefore since $1 < 3$ and $<$ is transitive:
$$x < 3$$
Now, $2 < 3$ therefore (invalid deduction) $2$ is a solution to the original inequality (false conclusion). And this is indeed a mistake people sometimes make when solving inequalities.
Or consider:
$$x^2 = -1$$
therefore
$$x^4 = 1$$
Now, in the integers or real numbers this also implies $x = 1 \vee x = -1$. But actually there are no solutions for $x^2 = -1$ in the real numbers. We can (in standard logic) actually deduce anything we like from $x \in \mathbb{R} \wedge x^2 = -1$, because it's false. One-way implications do not necessarily lead to solutions.
The problem stems from being taught various mechanisms for manipulating (in)equalities, but losing track of what these manipulations actually mean in terms of correct deductions about the values involved. If the operation applied to both sides is an injection then of course it's reversible and so we have two-way implication, and in that case the deduction would be valid.
In short: you can square both sides, of course. This tells you a true fact about $x$, but not everything true of $x$ in the squared equation is necessarily true of $x$ in the original. In this case $x = 1 \Rightarrow x \in \{1, -1\}$ but $x \in \{1, -1\} \not \Rightarrow x = 1$