What is X+Y=?

Noting that $11\leq XX \leq 99$ and the number with 4 digits will appear to right hand side for $XX$ with range of $31 < XX \leq 99\implies 33\leq XX\leq 99$

Let's make another notice that for $XX$ with in the range of $33^{Y}\leq XX^{Y} \leq 99^{Y}$ has exactly 4 digits for $Y = 2$ only . If $Y >2$ then $XX^{Y}$ will have digits greater than $4$ .

So we can draw out that $\begin{aligned} & XX^{Y} = XYYX \\& (10X+X)^Y = XYYX \\& (11X)^{Y} =10^3X+10^2Y+10Y+X\\& (11X)^Y = 10^3X+10Y\times 11 + X\end{aligned}$ For $2<x<9 , Y = 2$
$\begin{aligned} & (11X)^{2} = 10^3 X + 220 \end{aligned}$ Therefore, $\begin{aligned} & (11X)^{2} \neq 1001 X + 220 \end{aligned}$

Hence there exist no integer between $2< x<9$ to appear the number as per the problem . And only left integer for $X=1$ and also $\begin{aligned}& (11)^{Y} = 1001 + 110Y \\& \end{aligned}$ for which the value of $2<Y<4 \implies Y =3$ Hence $\begin{aligned}11^3 = 1001+330 = 1331 \end{aligned}$ $X+Y = 4$