What you need is the expansion of $(a+b)^4$. This is a special case of the very important Binomial Theorem. We have
This simplifies to $a^4+4a^3b+6a^2b^2+4ab^3+b^4$.
Put $a=y$, $b=1$, then $a=y$, $b=-1$, and add. There is a pleasant amount of cancellation.
Remarks: $1.$ The symmetrizing move $y=x+4$ is a useful idea.
$2$. For reasons of familiarity, we change the name, and study $(x+1)^4+(x-1)^4$. This function is symmetric about $x=0$. Our function is not $16$ at $x=0$, and by symmetry there are just as many solutions of $(x+1)^4+(x-1)^4=16$ with $x\gt 0$ as there are with $x\lt 0$. So let's see how many there are with $x\gt 0$.
$3.$ The solution $x=1$ is obvious. It is reasonably clear that there are no solutions with $0\lt x\lt 1$. And past $x=1$, our function is increasing. So there is exactly one positive solution, and therefore altogether there are two solutions.