At first, we see that $(a,b)=(9,1)$ works: Let $T$ denote the set of triangular numbers. If $t \in T$, let $t=\frac{n(n+1)}2$ and $m=3n+1$. Then $9t+1=\frac{9n^2+9n+2}2 = \frac{(3n+1)(3n+2)}2=\frac{m(m+1)}2 \in T$. Conversely, if $9t+1=\frac{m(m+1)}2$, we look $\mod 3$ to get that $m \equiv 1 \mod 3$, thus there is an integer $n$ with $m=3n+1$. The same identity like before shows that $t=\frac{n(n+1)}2 \in T$. So we proved $t\in T \iff 9t+1 \in T$. To get infinitely many pairs $(a,b)$, we just see them as linear maps and convolute them. Thus if $t \in T \iff at+b \in T$ and $t \in T \iff ct+d \in T$, we also have $t \in T \iff a(ct+d)+b=act+(ad+b) \in T$. It's clear that our pair $(9,1)$ generates infinitely many pairs because the first paramter $a$ always increases by this convolution.

Peter wrote: Prove that there exist an infinite number of ordered pairs $ (a,b)$ of integers such that for every positive integer $ t$, the number $ at+b$ is a triangular number if and only if $ t$ is a triangular number. Let $ \mathcal{T}$ be the set of triangular numbers: \[ \mathcal{T}=\left\{\frac{n(n+1)}{2}\;\vert\; n\in\mathbb{N}_{\geq 0}\right\}.\] We prove the following proposition. Suppose that $ d$ is an odd integer (so that $ d^{2}\equiv 1\; (mod\; 8)$). Let $ t$ be a positive integer. Then, $ t\in\mathcal{T}$ if and only if $ d^{2}t+\frac{d^{2}-1}{8}\in\mathcal{T}$. {Proof} $ (\Rightarrow)$ Assume that $ t\in\mathcal{T}$ so that $ t =\frac{\alpha(\alpha+1)}{2}$ for some $ \alpha\in\mathbb{N}_{\geq 0}$. Set $ l =\frac{d (2\alpha+1)-1}{2}$. Since $ d$ is odd, we see that $ l\in\mathbb{N}_{\geq 0}$. One may easily check that \[ d^{2}t+\frac{d^{2}-1}{8}=d^{2}\frac{\alpha(\alpha+1)}{2}+\frac{d^{2}-1}{8}=\frac{l(l+1)}{2}\in\mathcal{T}.\] $ (\Leftarrow)$ Assume that $ d^{2}t+\frac{d^{2}-1}{8}\in\mathcal{T}$ so that \[ d^{2}t+\frac{d^{2}-1}{8}=\frac{l(l+1)}{2}\] for some $ l\in\mathbb{N}_{\geq 0}$. It follows that \[ (8t+1) d^{2}=(2l+1)^{2}\] so that $ 8t+1$ is a square of an odd integer $ 2\alpha+1$, where $ \alpha\in\mathbb{N}_{\geq 0}$. Therefore, $ t =\frac{\alpha(\alpha+1)}{2}\in\mathcal{T}$.

Let $ a = (2l+1)^2, b= \frac{l(l+1)}{2}$ Whenever $at + b$ is triangular, say $k(k+1)/2$, By trivial calculations, $t = \frac{(\frac{k-l}{2l+1})(\frac{k-l}{2l+1} + 1)}{2}$ Since $t$ is an integer, we know that $\frac{k-l}{2l+1}$ is also an integer. And hence, $t$ is triangular. Similarly in the reverse case, we may express any integer $ m$ as $\frac{k-l}{2l+1}$ and hence the result follows.