SAMC

It is clear that $n=0$ and $n=1$ are solutions. Let $9n+16=x^2$, $16n+9=y^2$. Then $$16x^2-9y^2=16^2-9^2$$ that is $$(4x-3y)(4x+3y)=7\cdot 5^2.$$ We can assume that $x,y>0$, hence we have $4x-3y>0$. Considering the following possibilities $$\{\begin{array}{l}4x-3y=1\\ 4x+3y=175\end{array}\{\begin{array}{l}4x-3y=5\\ 4x+3y=35\end{array}\{\begin{array}{l}4x-3y=7\\ 4x+3y=25\end{array}$$ we get the solutions $x=22,5,8$, respectively. Hence $n=52$, $n=1$, and $n=0$ are the desired values of $n$.

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Alternative solution.

Let $9n+16=x^2$, $16n+9=y^2$. Then $$(xy{)}^2=(9n+16)(16n+9)=(12n{)}^2+337n+12^2,$$ hence $$(12n+15{)}^2>(xy{)}^2\ge (12n+12{)}^2$$ It follows that there are only three possibilities for $(xy{)}^2$, that is $(12n+12{)}^2$, $(12n+13{)}^2$, and $(12n+14{)}^2$. Solving the corresponding equation we get $n=0,1$ and $n=52$.