Solution:
In order for the expression $2^{x} + 3^{y}$ to be a perfect square, a positive integer $t$ such that $2^{x} + 3^{y} = t^{2}$ should exist.

Case 1. If $x$ is even, then there exists a positive integer $z$ such that $x = 2z$. Then
$$(t - 2^{z})(t + 2^{z}) = 3^{y}$$
Since $t + 2^{z} - (t - 2^{z}) = 2^{z+1}$, which implies $\gcd(t - 2^{z}, t + 2^{z}) \mid 2^{z+1}$, it follows that $\gcd(t - 2^{z}, t + 2^{z}) = 1$, hence $t - 2^{z} = 1$ and $t + 2^{z} = 3^{y}$, so we have $2^{z+1} + 1 = 3^{y}$.
For $z = 1$ we have $5 = 3^{y}$ which clearly has no solution. For $z \geq 2$ we have (modulo 4) that $y$ is even. Let $y = 2k$. Then $2^{z+1} = (3^{k} - 1)(3^{k} + 1)$ which is possible only when $3^{k} - 1 = 2$, i.e. $k = 1, y = 2$, which implies that $t = 5$. So the pair $(4, 2)$ is a solution to our problem.

Case 2. If $y$ is even, then there exists a positive integer $w$ such that $y = 2w$, and
$$(t - 3^{w})(t + 3^{w}) = 2^{x}$$
Since $t + 3^{w} - (t - 3^{w}) = 2 \cdot 3^{w}$, we have $\gcd(t - 3^{w}, t + 3^{w}) \mid 2 \cdot 3^{w}$, which means that $\gcd(t - 3^{w}, t + 3^{w}) = 2$. Hence $t - 3^{w} = 2$ and $t + 3^{w} = 2^{x-1}$. So we have
$$2 \cdot 3^{w} + 2 = 2^{x-1} \Rightarrow 3^{w} + 1 = 2^{x-2}$$
Here we see modulo 3 that $x - 2$ is even. Let $x - 2 = 2m$, then $3^{w} = (2^{m} - 1)(2^{m} + 1)$, whence $m = 1$ since $\gcd(2^{m} - 1, 2^{m} + 1) = 1$. So we arrive again to the solution $(4, 2)$.

Case 3. Let $x$ and $y$ be odd. For $x \geq 3$ we have $2^{x} + 3^{y} \equiv 3 \pmod{4}$ while $t^{2} \equiv 0, 1 \pmod{4}$, a contradiction. For $x = 1$ we have $2 + 3^{y} = t^{2}$. For $y \geq 2$ we have $2 + 3^{y} \equiv 2 \pmod{9}$ while $t^{2} \equiv 0, 1, 4, 7 \pmod{9}$. For $y = 1$ we have $5 = 2 + 3 = t^{2}$, clearly this doesn't have solution.

Note. The proposer's solution used Zsigmondy's theorem in the final steps of cases 1 and 2.