For which positive integers $n$ are there positive integers $x, y, z$ such that $4xyz=n(x+y+z)$?
An equivalence which will turn out useful later: A positive integer $n$ is the hypotenuse of a primitive Pythagorean triangle iff every factor of $n$ is congruent to 1 modulo 4.
Here's what I have so far. There is at least one solution $(x, y, z)$ for $n$ provided $n\ne 1\mod 12$:
$$\begin{align*} n &= 2k: (k+1, k, 1) \\ n &= 3k-1: (n, k, 1) \\ n &= 4k-1: ((k+1)n, k, 1) \\ n &= 12k-3: (4k^2+11k-3, k, 3) \end{align*}$$
Moreover, if $(x, y, z)$ solves $n=n_0$, then $(ax, ay, az)$ solves $n=a^2 n_0$. So we need only consider $n=p^2$ for prime $p$, and square-free $n$.
$n$ has a solution with $x=n$ if $n$ has the form $(4z-1)y-z$, so $4n+1=(4y-1)(4z-1)$. So if $4n+1$ is the product of two integers congruent to 3 modulo 4, then this gives such a solution. This fails only if $4n+1$ is the product of primes all of which are congruent to 1 modulo 4, i.e. $4n+1$ is the hypotenuse of a primitive Pythagorean triangle.
$n$ has a solution with $z=1$ if $n$ has a factor congruent to 3 modulo 4: Suppose $n=ab$, and $a=4d-1$. Then we have a solution with $x=a(y+1)$, $y=bd$, $z=1$: $$\begin{align*} 4d &= a+1\\ \implies 4(bd+1)d &= a(bd+1)+bd+1\\ \implies 4a(bd+1)bd &= ab[a(bd+1)+bd+1]\\ \implies 4xyz &= n(x+y+z) \end{align*}$$
Thus if $n$ is solutionless, every factor is congruent to 1 modulo 4, i.e. $n$ is the hypotenuse of a primitive Pythagorean triangle.
The lowest solutionless $n$ are 1, 13, $25=5^2$, 37, 97, $169=13^2$, 181, 193, $289=17^2$, 541, $625=5^4$, 673, 757, $841=29^2$, $949=13\cdot73$. Neither this list nor the list of primes in it is in OEIS. However, the list of solutionless squares is identifiable. In his answer, Denis Shatrov proved that $n=h^2$ is solutionless iff every factor of $h$ is congruent to 1 modulo 4.
I'll prove your conjecture that $n = h^2$ is solutionless if $h$ is the hypotenuse of a primitive Pythagorean triangle. That is, every prime divisor of $h$ is congruent to 1 modulo 4.
$$4xyz = h^2(x + y + z)$$ $$z(4xy - h^2) = h^2(x + y)$$
$4xy - h^2 \equiv 3 \pmod{4}$ since $h^2 \equiv 1 \pmod{4}$. Therefore $p \mid 4xy - h^2$ for a prime $p$ such that $p \equiv 3 \pmod{4}$. Also $p \mid h^2(x + y)$. Therefore $p \mid x + y$ ($x \equiv -y \mod{p}$).
$$4xy - h^2 \equiv -4y^2 - h^2 \equiv 0 \pmod{p}$$ $$h^2 \equiv -(2y)^2 \pmod{p}$$ Which is impossible, since $-1$ is a quadratic nonresidue modulo $p \equiv 3 \pmod{4}$.