Consider the equation: $ b^2+1=5a(a+1)$ $ \Leftrightarrow 4(b^2+1)=5(2a+1)^2-5$ $ \Leftrightarrow 4b^2-5(2a+1)^2=-9$ Let $ 2b=3x,2a+1=3y$ Then: $ x^2-5b^2=-1$ It is the Pell's equation:A root sequence: $ x_1=2,y_1=1$ $ x_2=38,y_2=17$ $ x_{n+1}=9x_n-x_{n-1}$ $ y_{n+1}=9y_n-y_{n-1}$ $ x_n\equiv 0(\mod 2),\forall n\in N$ $ \{y_n (\mod 2)\}$ is period $ y_{n}=y_{n+6}$ so it contain infinite odd number. Now con sider : $ b_n=\frac{3x_n}{2},a_n=\frac{3y_n-1}{2}$ Easy to check they satisfty the relative.

For reference: another solution can be found here.

another solution: Lemma:if $ n$ is not a multiple of $ 4$,and every odd prime divisor of $ n$ have form $ 4k + 1$,then there exist infinaitly many $ x$ such that $ n|x^2 + 1$ Now suppose that $ c_n$ be strictly increasing sequence such that for all natural $ n$:every prime divisor of $ c_n$ have the form $ 4k + 1$. let $ a_n = c_n^2$...therefore $ a_n(a_n + 1)$ is not a multiple of $ 4$ and every odd prime divisor of $ a_n(a_n + 1)$ have form $ 4k + 1$.(why?) therefore by Lemma,exist infinitly many $ x$ such that $ a_n(a_n + 1)|x^2 + 1$ now just define $ b_n =$ arbitary natural number $ x$,such that $ a_n(a_n + 1)|x^2 + 1$ and $ x > b_k$ for $ k < n$...