Find all pairs of positive integers $(x, y)$ such that $2^{x}+3^{y}$ is a perfect square.
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 $x=2 z$. Then

$$
\left(t-2^{z}\right)\left(t+2^{z}\right)=3^{y}
$$

Since $t+2^{z}-\left(t-2^{z}\right)=2^{z+1}$, which implies $g c d\left(t-2^{z}, t+2^{z}\right) \mid 2^{z+1}$, it follows that $\operatorname{gcd}\left(t-2^{z}, t+\right.$ $\left.2^{z}\right)=1$, 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 have no solution. For $z \geq 2$ we have (modulo 4) that $y$ is even. Let $y=2 k$. Then $2^{z+1}=\left(3^{k}-1\right)\left(3^{k}+1\right)$ 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 $y=2 w$, and

$$
\left(t-3^{w}\right)\left(t+3^{w}\right)=2^{x}
$$

Since $t+3^{w}-\left(t-3^{w}\right)=2 \cdot 3^{w}$, we have $\operatorname{gcd}\left(t-2^{z}, t+2^{z}\right) \mid 2 \cdot 3^{w}$, which means that $g c d(t-$ $\left.3^{w}, t+3^{w}\right)=2$. 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 $x-2=2 m$, then $3^{w}=\left(2^{m}-1\right)\left(2^{m}+1\right)$, whence $m=1$ since $\operatorname{gcd}\left(2^{m}-1,2^{m}+1\right)=1$. So we arrive again to the solution $(4,2)$.

Case 3. Let $x$ and $y$ be odd. For $x \geq 3$ we have $2^{x}+3^{y} \equiv 3(\bmod 4)$ while $t^{2} \equiv 0,1(\bmod 4)$, a contradiction. For $x=1$ we have $2+3^{y}=t^{2}$. For $y \geq 2$ we have $2+3^{y} \equiv 2(\bmod 9)$ while $t^{2} \equiv 0,1,4,7(\bmod 9)$. 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.