$$y^3 = z^4 - nx^2 = (z^2 - \sqrt{nx})(z^2 + \sqrt{nx}). \quad (1)$$
For $y = s^2 - nt^2$ with $(n, t) = 1$, we have
$$\begin{aligned} y^3 &= (s^3 - 3\sqrt{n}s^2t + 3nst^2 - n\sqrt{nt^3})(s^3 + 3\sqrt{n}s^2t + 3nst^2 + n\sqrt{nt^3}) \\ &= (z^2 - \sqrt{nx})(z^2 + \sqrt{nx}), \end{aligned}$$
so any $(x, z)$ satisfying
$$z^2 = s^3 + 3nst^2, \quad (2)$$
$$x = 3s^2t + nt^3 \quad (3)$$
satisfies the equation (1). One can always find $x$ satisfying the equation (3), so let us show that there are infinitely many $z$ satisfying equation (2). Let us divide the problem into two cases:

1. If $3n$ is not a square number, put $s = 1$. Then equation (2) becomes a Pell's equation
$$z^2 - 3nt^2 = 1. \quad (4)$$
We know that there are infinitely many solutions of Pell's equation, and if $(z_1, t_1)$ is the pair of smallest positive integers which is a solution of the equation, each $(k$-th) positive integral solution $(z_k, t_k)$ of the equation satisfies
$$z_k + \sqrt{3nt_k} = (z_1 + \sqrt{3nt_1})^k. \quad (5)$$
Now let us prove that there are infinitely many $(z_k, t_k)$ such that
$$x_k = 3t_k + nt_k^3, \quad y_k = 1 - nt_k^2, \quad z_k = (1 + 3nt_k^2)$$
are relatively prime to each other.
$$(x_k, z_k) \quad |(x_k, z_k^2) = (t_k(3 + nt_k^2), 1 + 3nt_k^2) \\ = (3 + nt_k^2, 1 + 3nt_k^2) (\because (t_k, 1 + 3nt_k^2) = 1) = (3 + nt_k^2, -8) = 1 \text{ or } 2,$$
$$(y_k, z_k) \quad |(y_k, z_k^2) = (1 - nt_k^2, 1 + 3nt_k^2) = (1 - nt_k^2, 4) = 1 \text{ or } 2,$$
$$(x_k, y_k) = (3t_k + nt_k^3, 1 - nt_k^2) = (4t_k, 1 - nt_k^2) = (4, 1 - nt_k^2) = 1 \text{ or } 2 \\ (\because (t_k, 1 - nt_k^2) = 1).$$
Therefore, if $n$ is even or $t_k$ is even, we have $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$. In the equation (5), one can easily check that if $k$ is even, $t_k$ is even. Therefore, there are infinitely many $(z_k, t_k)$ which gives us the solution of $nx^2 + y^3 = z^4$ and $(x, y) = (y, z) = (z, x) = 1$.

2. If $3n = m^2$ for some positive integer $m$, put $s = u^2, z = uv$. Then the equation (2) becomes
$$u^4 = v^2 - 3nt^2 = (v - mt)(v + mt). \qquad (6)$$
One can easily check that for each positive integer $k$,
$$\left( u_k = 2mk + 1, t_k = \frac{u_k^4 - 1}{m}, v_k = mt_k + 1 \right)$$
is a triple of integers satisfying the equation (6). Note that $u_k$ is odd, $t_k$ is even, and
$$(u_k, nt_k) \quad |(u_k, 3nt_k) = (u_k, m u_k^4 - m) = (u_k, m) = 1,$$
$$\therefore (u_k, nt_k) = 1,$$
$$(t_k, v_k) = (t_k, mt_k + 1) = 1.$$
Now the triple
$$(x_k = 3u_k^4 t_k + nt_k^3, y_k = u_k^4 - nt_k^2, z_k = u_k v_k = u_k(mt_k + 1))$$
satisfies the equation (1). We have to show $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$.
$$(x_k, y_k) = (3u_k^4 t_k + nt_k^3, u_k^4 - nt_k^2) = (4u_k^4 t_k, u_k^4 - nt_k^2) = (4, u_k^4 - nt_k^2) = 1 \\ (\therefore (u_k^4, u_k^4 - nt_k^2) = (t_k, u_k^4 - nt_k^2) = 1, u_k \text{ is odd}, t_k \text{ is even})$$
$$\begin{aligned} (y_k, z_k) &= (u_k^4 - nt_k^2, u_k v_k) = (u_k^4 - nt_k^2, v_k) | (u_k^4 - nt_k^2, v_k^2 = u_k^4 + 3nt_k^2) \\ &= (u_k^4 - nt_k^2, 4nt_k^2) = (u_k^4 - nt_k^2, 4) = 1, \\ (\therefore (nt_k^2, u_k^4 - nt_k^2) = 1, u_k \text{ is odd}, t_k \text{ is even}) \end{aligned}$$
$$\begin{aligned} (z_k, x_k) &= (3u_k^4 t_k + nt_k^3, u_k v_k) = (3u_k^4 + nt_k^2, u_k v_k) \\ &= (3u_k^4 + nt_k^2, v_k) | (3u_k^4 + nt_k^2, v_k^2 = u_k^4 + 3nt_k^2) \\ &= (-8nt_k^2, u_k^4 + 3nt_k^2) = (8, u_k^4 + 3nt_k^2) = 1 \\ (\therefore (nt_k^2, u_k^4 + 3nt_k^2) = 1, u_k \text{ is odd}, t_k \text{ is even}). \end{aligned}$$
So we have $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$. Therefore we can find infinitely many triples of integers $(x_k, y_k, z_k)$ with $(x_k, y_k) = (y_k, z_k) = (z_k, x_k) = 1$ which satisfies the equation (1).

From above, in any cases, we can find infinitely many integral solutions of the equation $nx^2 + y^3 = z^4$, $(x, y) = (y, z) = (z, x) = 1$. This completes the proof. □