Solving $(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$ with $x\in(-1,\infty)$ I've been struggling for a few hours on the below pre-calculus olympiad equation to which I still don't have an answer:

$$(x^2+4x+3)^x+(2x+4)^x=(x^2+4x+5)^x$$
  where $x \in (-1,\infty)$.

Now, according to WolframAlpha, this has a unique solution, $x=2$, and I need to prove this for one of my student without using derivatives or any more advanced techniques.
Interesting substitutions that I've tried, but with no success:
1) $a=(x+2)-\frac{1}{x+2}, b=(x+2)-\frac{1}{x+2}$ leads to 
$$(b^2-a^2)^x=(b^x-a^x)^2$$
2) $\alpha =2\arctan(x+2)$ leads to 
$$(-\cos(\alpha ))^{\tan(\alpha /2)-2} + (\sin(\alpha ))^{\tan(\alpha /2)-2} = 1$$
Both of them make the solution $x=2$ easy to see, but its unicity still eludes me.

First, the trivial solution is $x=2$.
Then, consider the fact $$(x^2+4x+3)^2+(2x+4)^2=(x^2+4x+5)^2 \tag{1}$$
And the fact $\dfrac{x^2+4x+3}{x^2+4x+5}$ and $\dfrac{2x+4}{x^2+4x+5}$ are both in $(0,1)$ if $x \in (-1,\infty)$. (I believe you can figure this out yourself.)
Consider that $a^x$ is strictly decreasing if $a\in (0,1)$.
We can easily get 
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x \gt \left({x^2+4x+3\over x^2+4x+5}\right)^2 \tag{2}$$ if $x\in (-1,2)$, and other cases.
We have:
Case:$x\in (-1,2)$, then 
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x+\left({2x+4\over x^2+4x+5}\right)^x \gt \left({x^2+4x+3\over x^2+4x+5}\right)^2+\left({2x+4\over x^2+4x+5}\right)^2=1 \tag{3}$$.
In another case ($x\in (2, \infty)$), we have
$$\left({x^2+4x+3\over x^2+4x+5}\right)^x+\left({2x+4\over x^2+4x+5}\right)^x \lt \left({x^2+4x+3\over x^2+4x+5}\right)^2+\left({2x+4\over x^2+4x+5}\right)^2=1 \tag{4}$$
Hence, the only solution is $x=2$.
