jmc

We can write the equation as $$x^4+y^4+z^4+1=4xyz.$$Then $xyz$ must be positive.

Let $a=|x|,$ $b=|y|,$ and $c=|z|,$ so $abc=|xyz|=xyz.$ Hence, $$a^4+b^4+c^4+1=4abc.$$By AM-GM, $$a^4+b^4+c^4+1\ge 4\sqrt[4]{a^4{b}^4{c}^4}=4abc.$$Since we have the equality case, we must have $a=b=c=1.$ Hence $|x|=|y|=|z|=1.$

Since $xyz$ is positive, the only possible triples are $(1,1,1),$ $(1,-1,-1),$ $(-1,1,-1),$ and $(-1,-1,1),$ giving us $\boxed{4}$ solutions.