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

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2. Solution: $\because x^{2} \equiv 0,1,4$ (mod5), $2^{2 k+1} \equiv 4^{k} \times 2$ $\equiv(-1)^{k} \times 2 \equiv 2$ or $3(\bmod 5)$,
$\therefore y$ must be even, let $y=2 z$, then $\left(2^{x}+x\right)\left(2^{x}-x\right)=615=3 \times 5 \times 41$, $\left\{\begin{array}{l}2^{x}+x=615, \\ 2^{x}-x=1,\end{array}\right.$ or $\left\{\begin{array}{l}2^{x}+x=205, \\ 2^{x}-x=3,\end{array}\right.$ or $\left\{\begin{array}{l}2^{x}+x=41, \\ 2^{z}-x=15,\end{array}\right.$ or $\left\{\begin{array}{l}2^{x}+x=123, \\ 2^{x}-x=5 .\end{array}\right.$ Solving the 4th equation yields the unique solution $z=6$. $\therefore x=59, y=12$.