Example 2 Find all positive integers $x, y$ such that $x^{2}+615=2^{y}$.

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Solution: For non-negative integer $k$, we have
$$
2^{2 k+1}=4^{k} \times 2 \equiv(-1)^{k} \times 2 \equiv 2 \text { or } 3(\bmod 5) \text {. }
$$

Since $x^{2}=0$ or 1 or $4(\bmod 5)$, $y$ must be even.
Let $y=2 z$, substituting into the given equation we get
$$
\left(2^{3}-x\right)\left(2^{5}+x\right)=615=3 \times 5 \times 41 \text {. }
$$

Thus, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ 2 ^ { z } + x = 6 1 5 , } \\
{ 2 ^ { z } - x = 1 }
\end{array} \text { or } \left\{\begin{array}{l}
2^{z}+x=205, \\
2^{z}-x=3
\end{array}\right.\right. \\
\text { or }\left\{\begin{array} { l } 
{ 2 ^ { z } + x = 4 1 , } \\
{ 2 ^ { z } - x = 1 5 }
\end{array} \text { or } \left\{\begin{array}{l}
2^{z}+x=123, \\
2^{z}-x=5 .
\end{array}\right.\right.
\end{array}
$$

Clearly, the first three systems of equations have no positive integer solutions. From the fourth system of equations, we get $2^{z}=64$, so $z=6$.
Thus, $x=59, y=12$.