$\textbf{Case 1: } p=q$ The set of the divisors of the number $n$ is: $D=\{1,p,p^2\}$. $1+p^2+p^4=p^4+6p^2+9\Longrightarrow 5p^2+8=0$, contradiction. Hence, we don't have natural solutions in this case. $\textbf{Case 2: } p,q$ are distinct The set of the divisors of the number $n$ is: $D=\{1,p,q,pq\}$. $1+p^2+q^2+p^2q^2=p^2q^2+6pq+9\Longrightarrow$ $\Longrightarrow \begin{cases}p^2+2pq+q^2=8pq+8;\\p^2-2pq+q^2=4pq+8.\end{cases}\Longrightarrow$ $\Longrightarrow \begin{cases}(p+q)^2=8pq+8;\\(p-q)^2=4pq+8.\end{cases}$ $\Longrightarrow \begin{cases}4\cdot2(n+1)=(p+q)^2;\\4(n+2)=(p-q)^2.\end{cases}$ The relations are possible only if $p,q$ are both odd numbers and results: $\begin{cases} 2(n+1)=\left(\dfrac{p+q}{2}\right)^2;\\n+2=\left(\dfrac{p-q}{2}\right)^2.\end{cases}$ With the notations $\dfrac{p+q}{2}=k;\dfrac{|p-q|}{2}=m$ results: $\begin{cases}2(n+1)=k^2;\\n+2=m^2;\end{cases}$ where $k,m\in\mathbb{N}$. The problem has solutions, there exist numbers with the given property. For example: $n=287=41\cdot7$.