Can we solve inequality without expanding both sides if each side has perfect square? The problem

$$\begin{align} (x+5)^2 &< (x-3)^2\tag{1} \end{align}$$

Normal people (most people) will solve the problem like this:
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
x^2 + 10x + 25 &< x^2-6x +9\\
10x+6x + 25-9 &< 0\\
16x &< -16\\
x &< -1
\end{align}$$
I want to know if I can solve $(1)$ like this:
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
(x+5)^2 - (x-3)^2 &<0\\
(2x+2)(7) &< 0 \tag{$a^2-b^2=(a+b)(a-b)$}\\
14x &< -14\\
x &< -1
\end{align}$$
Someone on facebook says like this:

You accidentally stumbled upon the right answer, but this process is
totally invalid and violates several properties of mathematics.

I want to know where the mistake is. Could you spot the mistakes please? Because I have tried on 3 different inequalities, and I still got the correct answers using this method. Thank you in advance!
P.S. edit version. I separate the edited version so that people can understand the problem and my mistakes.
$$\begin{align}
(x+5)^2 &< (x-3)^2\\
(x+5)^2 - (x-3)^2 &<0\\
(2x+2)(8) &< 0 \tag{$a^2-b^2=(a+b)(a-b)$}\\
16x &< -16\\
x &< -1
\end{align}$$

I see no big mistake and you were told wrong. The process is totally valid and who made that comment probably didn't bother to look at it (you just have a minor computational mistake).
The inequality $(x+5)^2<(x+3)^2$ is equivalent to
$$
(x+5)^2-(x-3)^2<0
$$
and the left-hand side can be factored to give
$$
(x+5-x+3)(x+5+x-3)<0
$$
hence
$$
8(2x+2)<0
$$
(you mistakenly wrote $7$). Now expanding the multiplication is possible, but one can better note that this can be rewritten as
$$
16(x+1)<0
$$
and the factor $16$ can be removed to give
$$
x+1<0
$$
and so
$$
x<-1
$$
