Let $ b=2y,a=3y+x$, then $ a^2+b^2-5-3ab=x^2-5y^2-5=0$ give solution. $ x+y\sqrt 5 =(5+2\sqrt 5)(9+4\sqrt 5 )^k$ is solution.

http://www.mathlinks.ro/Forum/viewtopic.php?t=170040 see more here.

Once you verified that for $a_0 = -1$, $a_1 = 1$, $a_{n+2} = 3a_{n+1} - a_n$ (careful, you have a typo there!) you have $a_n \mid a_{n+1}^2 - 5$ and $a_{n+1} \mid a_{n}^2 - 5$, it is clear we have $\gcd(a_n, a_{n+1}) \mid 5$. But modulo $5$ the sequence $(a_n)_{n\geq 0}$ is periodic, with period $-1,1$, and so no term is divisible by $5$, therefore $\gcd(a_n, a_{n+1}) = 1$. Also, by the Vieta jumping method, it is proved more than asked, namely that the solutions above are the only ones. One correction though; it is asked that both $\dfrac {a^2-5} {b}$ and $\dfrac {b^2-5} {a}$ be positive integers, for $a,b$ positive integers, thus $a,b > 2$, so the solutions start with $(4,11)\mapsto (11,29)\mapsto (29, 76)\mapsto \cdots$. Otherwise we may take terms $\pm a_n, \pm a_{n+1}$, and we also have the special solutions $(a,b) \in \{(1,2), (2,1)\}$, missing from your list, but verifying $ab \mid a^2 + b^2 - 5$.

If true, then $\tfrac{a^2-5}{b},\tfrac{b^2-5}{a}\in \bf Z$, so their product $\tfrac{(a^2-5)(b^2-5)}{ab}=\tfrac{a^2b^2-5a^2-5b^2+25}{ab}=ab-5\left(\tfrac{a^2+b^2-5}{ab}\right)$ is also integer. Hence, it suffices to show that there are infinitely many integers $a,b$ for which $ab\mid (a^2+b^2-5)$. Consider the infinite sequence: -1,1,4,11,29,76,199,..., defined by $a_0=-1$, $a_1=1$, and $a_{n+2}=3a_{n+1}-a_n\ \forall{n\ge{0}}$. We now quickly prove that $\gcd(a_n,a_{n+1})=1$ before moving onto the main proof. Note that we have $a_n \mid a_{n+1}^2 - 5$ and $a_{n+1} \mid a_{n}^2 - 5$, so it is clear that $\gcd(a_n, a_{n+1}) \mid 5$. But modulo $5$ the sequence $(a_n)_{n\geq 0}$ is periodic, with period $-1,1$, and so no term is divisible by $5$. Hence $\gcd(a_n, a_{n+1}) = 1$, as desired. I claim that $a_n^2+a_{n+1}^2=3a_na_{n+1}+5$. Proof: It is clearly true when $n=0$, for we get $a_0^2+a_1^2=(-1)^2+1^2=2=3(-1)(1)+5=3a_0a_1+5$. Suppose the proposition works for $n=k$, where $k\ge{0}$, i.e. $a_k^2+a_{k+1}^2=3a_ka_{k+1}+5$. By definition, $a_{k+2}=3a_{k+1}-a_k$. Adding $2a_{k+2}$ to both sides yields \[3a_{k+1}2a_{k+2}-a_k.\] Multiplying both sides by $a_{k+1}$ yields \[3a_{k+1}a_{k+2}=3a_{k+1}^2+2a_{k+1}a_{k+2}-a_ka_{k+1}.\] Adding 5 to both sides yields \begin{align*} 3a_{k+1}a_{k+2}+5 & =3a_{k+1}^2+2a_{k+1}a_{k+2}-a_ka_{k+1}+5\\ &=3a_{k+1}^2+2a_{k+1}a_{k+2}+3a_ka_{k+1}-4a_ka_{k+1}+5. \end{align*} By the induction hypothesis, $3a_ka_{k+1}+5=a_k^2+a_{k+1}^2$, thus \begin{align*} & 3a_{k+1}^2+2a_{k+1}a_{k+2}+3a_ka_{k+1}-4a_ka_{k+1}+5 \\ &=a_k^2+a_{k+1}^2+3a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}\\ &=a_k^2+a_{k+1}^2+3a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}\\ &=a_k^2+4a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}.\ (\clubsuit) \end{align*} But we can also get $(\clubsuit)$ as follows. By definition, $a_{k+2}=3a_{k+1}-a_k$. Subtracting $a_{k+1}$ from both sides, squaring, then adding $2a_ka_{k+1}$ to both sides yields \[a_k^2+a_{k+1}^2=a_k^2+4a_{k+1}^2+2a_{k+1}a_{k+2}-4a_ka_{k+1}.\] But this is just $(\clubsuit)$, hence \[\boxed{a_k^2+a_{k+1}^2=3a_ka_{k+1}+5}\] $\bf Q.E.D.$ Well, then, let $(a,b)=(a_{n+1},a_{n+2})=(a_{n+1},3a_{n+1}-a_n)$. (Do you see the Lemma coming?) Then, \begin{eqnarray*}\frac{a^2+b^2-5}{ab}&=&\frac{a_{n+1}^2+(3a_{n+1}-a_n)^2-5}{a_{n+1}(3a_{n+1}-a_n)}\\ &=&\frac{a_{n+1}^2+9a_{n+1}^2-6a_na_{n+1}+a_n^2-5}{a_{n+1}(3a_{n+1}-a_n)}\\ &=&\frac{3a_{n+1}(3a_{n+1}-a_n)+a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)}{a_{n+1}(3a_{n+1}-a_n)}. \end{eqnarray*} But by the Lemma, \[a_n^2+a_{n+1}^2=3a_na_{n+1}+5\Leftrightarrow a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)=0.\] Therefore, \begin{align*} & \frac{3a_{n+1}(3a_{n+1}-a_n)+a_n^2+a_{n+1}^2-(3a_na_{n+1}+5)}{a_{n+1}(3a_{n+1}-a_n)} \\ &=\frac{3a_{n+1}(3a_{n+1}-a_n)}{a_{n+1}(3a_{n+1}-a_n)}\\ &=\boxed{3}. \end{align*} Or, in other words, $\boxed{(a,b)=(a_{n+1},a_{n+2})}$ generates infinitely many pairs $(a,b)\in \bf Z$ for which $b\mid{a^2-5}$ and $a\mid{b^2-5}$, and we're done.

A simpler way to establish $a_{n}^2+a_{n+1}^2=3a_na_{n+1}+5$ where $a_0=(-1)$ , $a_1=1$ and $a_{n+2}=3a_{n+1}-a_n$ is : Assume true for n=k . Then , $(a_k+a_{k+1})^2=5(1+a_ka_{k+1})$ . We prove for $n=k+1$ as follows --- $(a_{k+2}+a_{k+1})^2=\left(5a_{k+1}-(a_k+a_{k+1})\right)^2=(a_k+a_{k+1})^2+25a_{k+1}^2-10a_{k+1}(a_k+a_{k+1})$ $=5(1+a_ka_{k+1})+15a_{k+1}^2-10a_ka_{k+1}=5+5a_{k+1}(3a_{k+1}-a_k)=5(1+a_{k+1}a_{k+2})$ This completes our inductive step in an easier way ,

Multiply the two fractions: $$\frac{(a^2 - 5)(b^2 - 5)}{ab} = ab + \frac{-5a^2 - 5b^2 + 25}{ab} = ab - 5\left(\frac{a^2 + b^2 + 5}{ab}\right)$$Now I claim that $\frac{a^2 + b^2 + 5}{ab} = 7$ has infinitely many positive integer solutions. Rearrange the equation and view it as a quadratic in terms of $a$: $$a^2 - 7ba + b^2 + 5 = 0$$The discriminant must be a perfect square so $49b^2 - 4(b^2 + 5) = k^2$ or $k^2 - 45b^2 = -20$ which has a solution $k = 5$, $b = 1$. And by well-known results, $k^2 - 45b^2 = 1$ has a solution, so there are infinitely many solutions to $k^2 - 45b^2 = -20$. If $b$ to be positive, $a$ must also be positive (otherwise, the left hand side of the quadratic would be positive). Thus, there are infinitely many $(a, b)$ such that both are positive integers that satisfy $\frac{a^2 + b^2 + 5}{ab} = 7$. Now restrict $b > 35$. Therefore, $ab > 35$ so $$\frac{(a^2 - 5)(b^2 - 5)}{ab} = an - 35$$is positive. Since $b > 35$, $b$ must $b^2 > 5$, so therefore $a^2 > 5$, otherwise, $ab - 35$ would not be positive.

My idea is sort of like what we do with vita jumping. let $(a,b)$ be a pair that have the property and $a>b$ now we shall assert $(a,\frac{a^2-5}{b})$ also have the property and we can also show that under some circumstances $\frac{a^2-5}{b}>a$, if $a > 5$: $\frac{a^2-5}{b}>a \iff \frac{a^2-5}{a}=a -\frac{5}{a}>a-1\ge b$ so if $\max(a,b)\ge 5$ then $\frac{a^2-5}{b}>a$ now suppose that $(a,b)$ have the property and now we show $(a,\frac{a^2-5}{b})$ have the property as well let $k=\frac{a^2-5}{b}$ then $\frac{k^2-5}{a} =\frac{(\frac{a^2-5}{b})^2-5}{a}=\frac{(a^2-5)^2-5b^2}{ab^2}$ and now it only take to prove: $a \mid (a^2-5)^2-5b^2 , b^2 \mid (a^2-5)^2-5b^2$ that both are straightforward to prove. we can also see $\frac{a^2-5}{k}=\frac{a^2-5}{\frac{a^2-5}{b}}=b \implies \frac{a^2-5}{k} \in N$ and the part is proving that $\gcd(k,a)=1$ we know that $\gcd(a,k)=\gcd(a,\frac{a^2-5}{b})$ now assume for some prime $p$: $p \mid a,p \mid \frac{a^2-5}{b} \implies p \mid a^2-5 \implies p \mid 5$ which a contradiction because $a\mid b^2 -5 \implies 5 \mid b$ so $\gcd(a,b) >5$ so we have proven that $(a,k)$ have the properties we need and also $a+k>a+b$ if $\max(a,b)>5$ now we put two initial values of $b=4,a=11$ and we see they have all the properties that we want and we can construct $k=29,a=11$ which has the property as well. and we know the method never get recurring because the sum of the new pair is always re-strictly greater than the sum of the previous pair.