Determine all positive integer numbers $k$ for which the numbers $k+9$ are perfect squares and the only prime factors of $k$ are 2 and 3.
We have an integer $x$ such that

$$
x^{2}=k+9
$$

$k=2^{a} 3^{b}, a, b \geq 0, a, b \in \mathbb{N}$.

Therefore,

$$
(x-3)(x+3)=k \text {. }
$$

If $b=0$ then we have $k=16$.

If $b>0$ then we have $3 \mid k+9$. Hence, $3 \mid x^{2}$ and $9 \mid k$.

Therefore, we have $b \geq 2$. Let $x=3 y$.

$$
(y-1)(y+1)=2^{a} 3^{b-2}
$$

If $a=0$ then $b=3$ and we have $k=27$.

If $a \geq 1$, then the numbers $y-1$ and $y+1$ are even. Therefore, we have $a \geq 2$, and

$$
\frac{y-1}{2} \cdot \frac{y+1}{2}=2^{a-2} 3^{b-2}
$$

Since the numbers $\frac{y-1}{2}, \frac{y+1}{2}$ are consecutive numbers, these numbers have to be powers of 2 and 3. Let $m=a-2, n=b-2$.

- If $2^{m}-3^{n}=1$ then we have $m \geq n$. For $n=0$ we have $m=1, a=3, b=2$ and $k=72$. For $n>0$ using $\bmod 3$ we have that $m$ is even number. Let $m=2 t$. Therefore,

$$
\left(2^{t}-1\right)\left(2^{t}+1\right)=3^{n}
$$

Hence, $t=1, m=2, n=1$ and $a=4, b=3, k=432$.

- If $3^{n}-2^{m}=1$, then $m>0$. For $m=1$ we have $n=1, a=3, b=3, k=216$. For $m>1$ using $\bmod 4$ we have that $n$ is even number. Let $n=2 t$.

$$
\left(3^{t}-1\right)\left(3^{t}+1\right)=2^{m}
$$

Therefore, $t=1, n=2, m=3, a=5, b=4, k=2592$.

Set of solutions: $\{16,27,72,216,432,2592\}$.