Solve over the integers

$\begin{aligned} 615 + x^2 & = 2^y \\ 2^y - x^2 & = 615 \\ \left(2^\frac y2 - x\right)\left(2^\frac y2 + x\right) & = 615 = \color{#3D99F6} 3\times 5 \times 41 & \small \color{#3D99F6} \text{3 prime factors} \end{aligned}$

Since we know that $2^\frac y2 - x < 2^\frac y2 + x$ , we can assume

$\begin{cases} 2^\frac y2 - x = 3\times 5 = 15 & ...(1) \\ 2^\frac y2 + x = 41 & ...(2) \end{cases}$

$\begin{aligned} (1)+(2): \quad 2 \times 2^\frac y2 & = 15+41 \\ \implies 2^\frac y2 & = \frac {56}2 & \small \color{#3D99F6} \text{No integer solution} \end{aligned}$

Similarly, there is no integer solution, if $2^\frac y2 - x = 3$ and $2^\frac y2 + x = 205$ . But $\begin{cases} 2^\frac y2 - x = 5 \\ 2^\frac y2 + x = 123 \end{cases} \implies 2^\frac y2 = \frac {128}2 = 64 \implies y = \boxed{12}$ and $x=\boxed {59}$ .

Since 615 + x^2 should be some power of 2 ,hence the only possibility of x and y 28 and 12 respectively.. As 615+3481(which is the square of 59) = 4096 which is 2 raise to the power 12 .