Needs a elegant solution apart from hit and trial (1)

$x+\sqrt{y}=27\Rightarrow x=27-\sqrt{y}...(1)$ $\sqrt{x}-y=1\\ \Rightarrow x=(y+1)^2$ $\Rightarrow 27-\sqrt{y}=y^2+2y+1~~(\text{Using (1)})$ Let $\sqrt {y}=t$ $\Rightarrow t^4-18t^2-t-54=0\\ \Rightarrow (t-2)(t^3+2t^2+6t+13)=0$ $\Rightarrow \sqrt {y} =t=2~\color{#302B94}{(**)}$ $\Rightarrow y=4 ,x=27-\sqrt{y}=25$ $\Large 25+4=\huge\boxed{\color{#007fff}{29}}$ .

$\color{forestgreen}{\boxed{\color{#302B94}{**\text{Further Analysis}}}}$

Let f(t)= $t^3+2t^2+6t+13$ . Differentiating $t^3+2t^2+6t+13$ we can easily see that it is a increasing function(and also: $\lim_{x\rightarrow\infty}f(t)=+\infty, \lim_{x\rightarrow{-\infty}}f(t)=-\infty$ which implies it can yield only one real t) and also f(-3)<0 and f(-2)>0 implies that only value of t lies between -3 and -2 {Apply Intermediate Value Theorem }which can be easily rejected since y is a positive integer.