Hint. Show the infinitude of positive integer solutions $(a,b)$ to the divisibility condition $ab\mid a^2+b^2-5$. In fact, for a positive integer $k$, there exists $(a,b)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ such that $$a^2+b^2-5=kab\tag{*}$$
if and only if $k=3$, in which case there are infinitely many choices of $(a,b)$. When $k=3$, amongst the positive integer solutions $(a,b)$ such that $a\geq b$, the smallest of which is $(a,b)=(4,1)$.
The idea is the technique known as Vieta jumping. If you do this correctly, then you will see that all positive integer solutions $(a,b)$ with $a\geq b$ to (*) with $k=3$ are of the form $(a,b)=(x_n,x_{n-1})$ for some positive integer $n$, where $(x_n)_{n=0}^\infty$ are given by $x_0=1$, $x_1=4$, and
$$x_n=3x_{n-1}-x_{n-2}$$
for every integer $n\geq 2$. Here is a closed form of $\left(x_n\right)_{n=0}^\infty$:
$$x_n=\left(\frac{1+\sqrt5}{2}\right)^{2n+1}+\left(\frac{1-\sqrt5}{2}\right)^{2n+1}=L_{2n+1}$$
for all $n=0,1,2,\ldots$, where $(L_r)_{r=0}^\infty$ is the sequence of Lucas numbers. The first few terms of $(x_n)_{n=0}^\infty$ are
$$1,4,11,29,76,199,521,1364,3571,9349,24476,\ldots\,.$$ Compare the list above with the answer by Arthur.