Proof

Let $n$ be a natural number equal the product of all its divisors $n=\prod_{d\mid n}d.$
While the index $d\mid n$ runs through all divisors of $n$, this holds also for the index $\frac nd\mid n$ built from the complementary divisor.
It follows $$n^4=n^2\cdot n^2=\prod_{d\mid n}d\cdot \prod_{d\mid n}\frac nd=\prod_{d\mid n}\left(d\cdot \frac nd\right)=\prod_{d\mid n}n=n^{\tau(n)},$$ with $\tau(n)$ being the number of positive divisors of $n.$
Taking the formula for calculating the number of positive divisors, the exponents in the factorization $n=p_1^{e_1}\cdots p_r^{e_r}$ obey the following restriction: $$(e_1+1)\cdots(e_r+1)=4.$$
This restriction can be only fulfilled, if $r=1$ and $e_i=3$ or if $r=2$ and $e_i=e_j=1$ for some $i\ge 1$, $j\ge 1$, with $i\neq j.$
From this, it follows that $n=p^3$ for some prime number $p$ (case $r=1$) or that $n=p_ip_j$ for some prime numbers $p_i\neq p_j$ (case $r=2$).

Let $n=p^3$ for be some prime number $p.$
- It follows $n=\prod_{d\mid p^2}d=1\cdot p\cdot p^2\cdot p^3=p^6=(p^3)^2=n^2.$
Let $n=p_ip_j$ for some prime numbers $p_i\neq p_j.$
- It follows $n=\prod_{d\mid p_ip_j}d=1\cdot p_i\cdot p_j\cdot p_ip_j=p^6=(p_ip_j)^2=n^2.$
  ∎

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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927