Can't prove this elementary algebra problem $x^2 + 8x + 16 - y^2$
First proof:
$(x^2 + 8x + 16) – y^2$
$(x + 4)^2 – y^2$
$[(x + 4) + y][(x + 4) – y]$
2nd proof where I mess up:
$(x^2 + 8x) + (16 - y^2)$
$x(x + 8) + (4 + y)(4 - y)$
$x + 1(x + 8)(4 + y)(4 - y)$ ????
I think I'm breaking one of algebra's golden rules, but I can't find it.

Let us fix your second approach. 
$$
\begin{align}
&x^2+8x+16-y^2\\
=&x^2+8x+(4-y)(4+y)\\
=&x^2+x\big[(4-y)+(4+y)\big]+(4-y)(4+y)\\
=&(x+4-y)(x+4+y).
\end{align}
$$
