Half way there

If the number we are looking for is $n$ and $m$ is the integer for which $2\times \sum_{i=1}^{m}i=m(m+1)=\frac{n(n+1)}{2}$ , then we can solve the equation for $m$

$m(m+1)=\frac{n(n+1)}{2} \implies m=\frac{-2\pm 2\sqrt{1+2n(n+1)}}{4}$

so we Need the Argument of the square root to be an integer $1+2n(n+1)=n^2+(n+1)^2=k^2$ for some postive integer $k$ . We Need to look for leg-leg Pythagorean triples then. the next one of them is $(119,120,169)$ , which makes $n=119$

We want to find positive integers $m,n$ such that $2 \times \tfrac12m(m+1) = \tfrac12n(n+1)$ , which simplifies to read $(2n+1)^2 - 2(2m+1)^2 \; = \; -1$ and so we need to look for solutions of Pell's equation. The solutions in positive integers of $x^2 - 2y^2 \;= \; -1$ and are given by $x_n + y_n \sqrt{2} \; = \; (1 + \sqrt{2})^{2n+1} \hspace{2cm} n \in \mathbb{N} \cup \{0\}$ We note that $x_2 + y_2\sqrt{2} = (1 + \sqrt{2})^5 \; = \; 41 + 29\sqrt{2}$ , which gives the $n=20$ , $m=14$ solution in the question. The next pair is $x_3 + y_3\sqrt{2} \; = \; (1 + \sqrt{2})^7 \; = \; 239 + 169\sqrt{2}$ which gives $n = \boxed{119}$ and $m=84$ as the next solution.