1. Simplify the differential equation

Given

$$\frac{dy}{dx} = \frac{2^{x+y}-2^x}{2^y}$$

Factor $2^x$ in the numerator:

$$\frac{dy}{dx} = \frac{2^x(2^y-1)}{2^y} = 2^x\left(1-\frac{1}{2^y}\right)$$

So,

$$\frac{dy}{dx} = 2^x\left(1-2^{-y}\right)$$

2. Separate variables

$$\frac{dy}{1-2^{-y}} = 2^x\,dx$$

Now simplify the left side:

$$1-2^{-y} = 1-\frac{1}{2^y} = \frac{2^y-1}{2^y}$$

Hence,

$$\frac{1}{1-2^{-y}} = \frac{2^y}{2^y-1}$$

Therefore,

$$\frac{2^y}{2^y-1}\,dy = 2^x\,dx$$

3. Integrate both sides

We need to evaluate

$$\int \frac{2^y}{2^y-1}\,dy$$

Let

$$t = 2^y \implies dt = 2^y\ln 2\,dy$$

so

$$dy = \frac{dt}{t\ln 2}$$

Then

$$\int \frac{2^y}{2^y-1}\,dy = \int \frac{t}{t-1}\cdot \frac{dt}{t\ln 2} = \frac{1}{\ln 2}\int \frac{dt}{t-1}$$

Thus,

$$\int \frac{2^y}{2^y-1}\,dy = \frac{1}{\ln 2}\ln|t-1|+C = \frac{1}{\ln 2}\ln(2^y-1)+C$$

Also,

$$\int 2^x\,dx = \frac{2^x}{\ln 2}+C$$

So the integrated equation is

$$\frac{1}{\ln 2}\ln(2^y-1) = \frac{2^x}{\ln 2}+C$$

Multiplying by $\ln 2$,

$$\ln(2^y-1) = 2^x + C_1$$

4. Use the initial condition $y(0)=1$

At $x=0$, $y=1$:

$$\ln(2^1-1) = 2^0 + C_1$$ $$\ln(1) = 1 + C_1$$ $$0 = 1 + C_1 \implies C_1 = -1$$

Hence,

$$\ln(2^y-1) = 2^x - 1$$

Exponentiating,

$$2^y - 1 = e^{2^x-1}$$ $$2^y = 1 + e^{2^x-1}$$

Therefore,

$$y = \log_2\left(1+e^{2^x-1}\right)$$

5. Find $y(1)$

Substitute $x=1$:

$$y(1)=\log_2\left(1+e^{2^1-1}\right)=\log_2(1+e)$$

6. Match with the options

$$y(1)=\log_2(1+e)$$

So the correct option is B.