Multiplying both sides by $1-x-x^{2}-x^{3}$ the right hand side becomes $a_{0}+\left(a_{1}-a_{0}\right) x+\left(a_{2}-a_{1}-a_{0}\right) x^{2}+\ldots$, and setting coefficients of $x^{n}$ equal to each other we find that $a_{0}=1, a_{1}=1, a_{2}=2$, and $a_{n}=a_{n-1}+a_{n-2}+a_{n-3}$ for $n \geq 3$. Thus the sequence of $a_{n}$ 's starts $1,1,2,4,7,13,24,44,81,149, \ldots$ So we now see that $a_{0}=1^{2}$ and $a_{8}=9^{2}$. What makes it impossible for this to happen again is that the sequence is growing exponentially. It will suffice to show that $a_{n}>1.5^{n}$ for $n>2$, since $n^{2} /(n-1)^{2}<1.5$ for $n \geq 6$, thus when $a_{n-1}$ exceeds $n^{2}$ at $n=10$ there can be no more solutions to $a_{n-1}=n^{2}$. Observe that $a_{n}>1.5 a_{n-1}$ for $n=3,4,5$. By way of induction, assume it for $n-2$, $n-1$, and $n$, then $a_{n+1}=a_{n}+a_{n-1}+a_{n-2}>1.5^{n}+1.5^{n-1}+1.5^{n-2}=1.5^{n-2}\left(1+1.5+1.5^{2}\right)>1.5^{n+1}$. Thus, by induction, $a_{n}>1.5^{n}$ for $n>2$, so the only solutions are $\mathbf{1}, \mathbf{9}$.