I think that if we choose $a_{k}=\frac{x_{2k+1}^{2}}{2}-1$ where $x_{2k+1}$ is the $2k+1-$th solution of Pell's equation $x^{2}-3y^{2}=1$ we are done...the first and the third conditions are satisfied but there are many computations about the second one or... I'm too blind.

A my friend said that : We can choose $m=588.$

orl wrote: Prove that there exists $m \in \mathbb{N}$ such that there exists an integral sequence $\lbrace a_{n}\rbrace$ which satisfies: I. $a_{0}= 1, a_{1}= 337$; II. $(a_{n+1}a_{n-1}-a_{n}^{2})+\frac{3}{4}(a_{n+1}+a_{n-1}-2a_{n}) = m, \forall$ $n \geq 1$; III. $\frac{1}{6}(a_{n}+1)(2a_{n}+1)$ is a perfect square $\forall$ $n \geq 1$. Let $b_{n}=4a_{n}+3$, then I)$b_{0}=7,b_{1}=1351$, II) $b_{n+1}b_{n-1}-b_{n}^{2}=16m$. Solution II) is $b_{n}=ax^{n}+by^{n}$, were x, y roots of $x^{2}-2cx+1=0, c\in Z/2.$ Therefore $x,y=c \pm \sqrt{c^{2}-1}$. Then $b_{n+1}b_{n-1}-b_{n}^{2}=ab(x^{2}+y^{2}-2)=4ab(c^{2}-1)=16m.$ It give $a,b=\frac{7}{2}\pm \frac{1351-7c}{2\sqrt{c^{2}-1}}, \ 8m=2ab(c^{2}-1)=9457c-912625.$ Therefore $c=1(mod \ 8)$. Therefore if c integer, then we satisfyed I) and II) and $b_{n+1}=cb_{n}-b_{n-1}$. Consider III). If $c_{n}^{2}=\frac{(a_{n}+1)(2a_{n}+1)}{6}$, then $(4a_{n}+3)^{2}-3(4b_{n})^{2}=1$ or $b_{n}^{2}-3(4c_{n})^{2}=1.$ It give $4a_{n}+3=b_{n}=\frac{(2+\sqrt 3)^{k(n)}+(2-\sqrt 3)^{k(n)}}{2}\ \forall n$. It give $k(0)=2, k(1)=6$. It may be if $c=\frac{(2+\sqrt 3)^{l}+(2-\sqrt 3 )^{l}}{2}, \ l\in N$. Because $c=1(mod \ 8)$ we get $l=4k$ or $c=\frac{(2+\sqrt 3)^{4k}+(2-\sqrt 3 )^{4k}}{2}.$ It give work $m=\frac{9457[(2+\sqrt 3)^{4k}+(2-\sqrt 3 )^{4k}]-1825250}{16}, \ k\in N.$

In fact I think the answer could only be 588~

I think that Rugini's solution is correct. Why only m=588 is a correct answer?