Since $2^x+17=y^4$ $\Longrightarrow$ $2^x\equiv y^4\pmod {17}$, if $x$ is odd $\Longrightarrow$ $2^x\equiv {2,8,9,15}\pmod {17}$, but $y^4\equiv {1,4,13,16}\pmod {17}$ which is a contradiction $\Longrightarrow$ $x$ is even. Let $x=2k$ $\Longrightarrow$ $17=y^4-2^{2k}$ $\Longrightarrow$ $17=(y^2-2^k)(y^2+2^k)$ since $17$ is prime we get $y^2-2^k=1$ and $y^2+2^k=17$ $\Longrightarrow$ $y=3$ and $k=3$ $\Longrightarrow$ all solutions $(x,y)=(6,3)$.

Note that $x>4$ so we have $ 2^4(2^{x-4} +1)= (y^2+1)(y^2-1)$ and $ gcd(y^2+1,y^2-1)=2$ . Let put $ y^2+ 1=2a$ and $ y^2-1=2b$ and $ gcd(a,b)=1$ .So we have $4(2^{x-4} +1)=ab$ , so $a=4$ or $b=4$ that gives $y=3 $ and $ x=6 $ So $ S=(6,3)$

I was same as FabrizioFelen thanks

FabrizioFelen wrote: if $x$ is odd $\Longrightarrow$ $2^x\equiv {2,8,9,15}\pmod {17}$, but $y^4\equiv {1,4,13,16}\pmod {17}$ which is a contradiction.

Notice $y^4\equiv \{0,1,4,13,16\}\pmod{17}$ and $2^x\equiv \{1,4,16,13\}$ if $x$ is even and $2^x\equiv \{2,8,15,9\}$ if $x$ is odd. Hence, $x$ is odd so let $x=2a.$ Then, $$(y^2-2^a)(y^2+2^a)=y^4-2^x=17.$$Then, $(y^2-2^a,y^2+2^a)=(1,17),(-1,17)$ since $2^a>0.$ We see $(y,a)=(3,3)$ is the only solution so $(x,y)=(6,3).$ $\square$

$\mod 128$ basically finishes it! For $x \le 6$, it is easy to see only $(x,y) = (6,3)$ works. Next we show $x \ge 7$ admits no solution. Assume contrary. Then, $$ 128 \mid y^4 - 17$$So clearly $y$ must be odd. Write $y = 8k + 1$. Then, \begin{align*} 0 \equiv y^4 - 17 = (8k+1)^2 - 17 &= (64k^2 + 16k + 1) - 17 = 16(4k^2 + k - 1) \pmod{128} \\ &\implies 4k^2 + k \equiv 1 \pmod{8} \end{align*}But by considering two cases according to whether $k$ is even or odd, it is easy to see that above is clearly false. $\blacksquare$

Taking the equation modulo $3$ implies that $x$ is even. Now if $x=2z$ : $$17 = (y^2+2^z)(y^2-2^z) \Rightarrow (x,y) = (6,3)$$ $$\mathbb{Q.E.D.}$$

Taking modulo $17$ tells us that $x$ needs to be even. Let $x = 2k$. We then have: $$(y^2+2^k)(y^2-2^k) = 17.$$From here we know that $y^2-2^k = 1$, so $y = 3$, $k=3$ is the only solution. Our pair $(x, y)$ is $\boxed{(6, 3)}$.

Only $(x,y)=(6,3)$ works. We may check if $x$ is odd then $2^x\equiv 2,8,-2,9\pmod 17$ and $y^4\equiv 1,13,4,-1$ none of these overlap so contradiction. Hence $x$ even so we may write $17=(y^2-2^\frac{x}{2})((y^2+2^\frac{x}{2}))$. Clearly $1=y^2+2^\frac{x}{2}$ is absurd, thus $1=y^2-2^\frac{x}{2}$. Rearranging $(y-1)(y+1)$ is a power of two, each must thus be a power of $2$, hence $2,4$ respectively. This gives the desired solution set, which we can check works.

Taking $\mod 5$ gives that $2^x+2 \equiv y^4 \equiv 1$ (by FLT) so $2^x=4$. This is only satisfied when $x \equiv 2 \pmod{4}$, and in particular, $x$ is even. Therefore, we can rewrite as $17=(2^x+y^2)(y^2-2^x)$ of which the only solution is $\boxed{(6,3)}$.

Bump Sol 2^x = y⁴ mod 17 So 2^4x = y^16 mod 17 So y^16 = 0 ,1 mod 17 Clearly 2^4x = 0 mod 17 has no sol So 2^4x = 1 mod 17 16^x = (-1)^x = 1 mod 17 So x is even Let x = 2k So 17= y⁴-2^2k = (y²-2^k)(y²+2^k) So , y² + 2^k = 17 y ² - 2^k = 1 2y² = 18 y = 3 2^k = 17 -9 2^k = 2³ k = 3 x = 6 So Only 1 solution (x,y ) = (6,3) Nice one