The answer are all odd squares. Note that $a$ is retriangular if $\exists b\in\mathbb{N}$ such that for any $n\in\mathbb{N}$, there is a $k_n\in\mathbb{N}$ so that \[ a\frac{n(n+1)}{2} + b = \frac{k_n(k_n+1)}{2}. \]This rearranges to \[ 4an(n+1) + 8b +1 = (2k_n+1)^2 \iff a(2n+1)^2 +8b+1-a = (2k_n+1)^2\triangleq y_n^2. \]Note that $(y_n)_{n\ge 1}$ is a strictly increasing sequence. Moreover, inspecting $y_{3n+1}^2$, we get \[ a(6n+3)^2 +8b+1-a = y_{3n+1}^2 \implies 9y_n^2-y_{n+1}^2 = (3y_n-y_{n+1})(3y_n+y_{n+1}) = 8(8b+1-a). \]From here, we deduce \[ 3y_n + y_{n+1}\mid \bigl|8b+1-a\bigr|. \]As $y_n$ is increasing, this yields $a=8b+1$. This, in return, leaves us with $a(2n+1)^2 = y_n^2$, where $y_n$ is odd. From here, we obtain that the answer are indeed all odd squares. Conversely if $a=(2k+1)^2 = 4k^2+4k+1$ then it suffices to choose $b=\frac{a-1}{8} = \frac{k(k+1)}{2}$.