For which $k \in \mathbb{N}$ do there exist non-negative integers $a$ and $b$ such that

$$
2^{a} 3^{b}=k(k+1) ?
$$
$\operatorname{pgcd}(k, k+1)=1$ so one factor is a power of 3 while the other is a power of 2.

Case $1: k$ is a power of 2 and $k+1$ is a power of 3, so $2^{a}+1=3^{b}$. Studying the cases $a=0, a=1$ provides us with the solution ( $a=1, b=1, k=2$ ). If $a \geq 2$, studying the equation modulo 4 ensures that $b$ is even. $b=2 b^{\prime}$, so $2^{a}=\left(3^{b^{\prime}}-1\right)\left(3^{b^{\prime}}+1\right)\left(3^{b^{\prime}}-1\right)$ and $\left(3^{b^{\prime}}+1\right)$ are two consecutive powers of 2. Therefore, $b^{\prime}=1$, leading to the solution $(a=3, b=2, k=8)$.

Case $2: k$ is a power of 3 and $k+1$ is a power of 2, so $2^{a}=3^{b}+1$. Studying the cases $a=0, a=1, a=2$ provides us with the solutions ( $a=1, b=0, k=1$ ) and ( $a=2, b=1, k=3$ ). If $a \geq 3$, studying the equation modulo 8 gives $3^{b}=-1[8]$, which has no solutions.

Therefore, the integers $k$ are $1, 2, 3$, or 8.