Following a hint by Mostafa Ayaz, we have $(n+1)(2n+1)=6k^2$ for some integer $k$. That is,
$$(4n+3)^2-3(4k)^2=1\,.$$
Hence, $(4n+3)+(4k)\sqrt{3}=(2+\sqrt{3})^m$ for some nonnegative integer $m$. Therefore,
$$4n+3=\sum_{r=0}^{\left\lfloor\frac{m}{2}\right\rfloor}\,\binom{m}{2r}\,2^{m-2r}\,3^r\,.$$
If $m$ is odd, then
$$4n+3\equiv 2m\cdot 3^{\frac{m-1}{2}}\pmod{4}\,.$$
If $m$ is even, then
$$4n+3\equiv 3^{\frac{m}{2}}\pmod{4}\,.$$
Since $$4n+3\equiv 3\pmod{4},$$ we need $$m\equiv 2\pmod{4}\,.$$
That is, $m=4s+2$ for some nonnegative integer $s$
$$4n+3+(4k)\sqrt{3}=(7+4\sqrt{3})\,(97+56\sqrt{3})^s\,.$$
That is, $n=a_s$ and $k=b_s$, where
$$a_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s+(7-4\sqrt{3})\,(97-56\sqrt{3})^s-6}{8}$$
and
$$b_s:=\frac{(7-4\sqrt{3})\,(97+56\sqrt{3})^s-(7-4\sqrt{3})\,(97-56\sqrt{3})^s}{8\sqrt{3}}\,.$$
Note that $a_0=1$, $a_1=337$, and
$$a_s=194\,a_{s-1}-a_{s-2}+144\text{ for }s=2,3,4,\ldots\,.$$
Furthermore, $b_0=1$, $b_1=195$, and
$$b_s=194\,b_{s-1}-b_{s-2}\text{ for }s=2,3,4,\ldots\,.$$
Therefore, the next smallest pair $(n,k)$ apart from the one given by Robert Israel is
$$(n,k)=(a_2,b_2)=(65521,37829)\,.$$