I could tell by eye that $z=2$ is a solution.
(Looking at divisors of $14$ would be a more formal approach to find that.)
which is $7$ when $z=0$ and strictly increasing for $z>0$,
so there are no other real positive solutions.
Correction in response to comment:
I showed that there are no other real positive solutions for $z$, but the question asked for real positive solutions for $y$, and there is another (irrational) one: $z^3+2z^2+4z+7 $ is negative $(1)$ when $z=-2$ and positive $(4)$ when $z=-1$, so there is a solution $z_0$ between $-2$ and $-1$, and $y_0=14+z_0$ is positive. Since $z^3+2z^2+4z+7$ is strictly increasing, there are no other real zeroes of $z^4-z-14$.