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

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