5. Let $\left\{\begin{array}{l}x^{2}+3 y=u^{2} \\ y^{2}+3 x=v^{2}\end{array}\right.$ Since $x, y$ are positive integers, we have $u>x, v>y$. We set $u=x+a, v=y+b$, where $a, b$ are positive integers. From
$$\left\{\begin{array}{l} x^{2}+3 y=(x+a)^{2} \\ y^{2}+3 x=(y+b)^{2} \end{array}\right.$$

we can derive
$$\left\{\begin{array}{l} 3 y=2 a x+a^{2} \\ 3 x=2 b y+b^{2} \end{array}\right.$$

Solving this system of linear equations in $x, y$ yields $\left\{\begin{array}{l}x=\frac{2 a^{2} b+3 b^{2}}{9-4 a b}, \\ y=\frac{2 b^{2} a+3 a^{2}}{9-4 a b} .\end{array}\right.$ Since $x, y$ are positive integers, we have $9-4 a b>0$. Since $a, b$ are positive integers, we have $a b=1$ or 2, i.e., $(a, b)=(1,1),(1,2)$, $(2,1)$. Correspondingly, we obtain $(x, y)=(1,1),(16,11),(11,16)$.