By the identity: $1^3 + 2^3 +\cdots + n^3 = \Big(\displaystyle\frac{n(n+1)}{2}\Big)^2$, we have:

$$\Big(\frac{(n-1)n}{2}\Big)^2 +n^3 = \Big(\frac{n(n+1)}{2}\Big)^2 $$

All we need to show is there are infinitely many positive integers $n$ such that $\displaystyle\frac{n(n+1)}{2}=k^2$ for some integers $k$. Then $(x, y, z)=\Big(\frac{n(n-1)}{2}, n, k\Big)$ solves the problem.

Now $\Big(\frac{n(n+1)}{2}\Big)=k^2$ can be rewritten as $(2n+1)^2 - 2(2k)^2 = 1$. It is clearly that $x^2-2y^2 = 1$ has infinitely many positive integer solutions. For any such $(x, y)$, $x$ is obviously odd. Let it be $x=2n+1$. Then $y^2=2n^2 + 2$ implies $y$ is even. Therefore $(x, y)$ must be of the form $(2n+1, 2k)$. As such, there are infinitely many such $n$.