I dont know whether i am anywhere close to the solution, but i got using modulo some powers of two that x is of the from 8l+2 whereas y is of the form 8l+1 Any help will be appreciated Thx, Befuddlers

After more than 2 years, I finally got a solution: Problem 17- A sum of prime powers is not a power of 2. Btw, I copied it below: Solution. We will prove that if $(x,y,z)$ is a solution that $(x,y,z)\in S:=\{(0,0,1), (0,1,3), (2,1,5)\}$. 1. If $x=y=0$ then $z=1$, so that $(0,0,1) \in S$. 2.If $x=0$ and $y\ge 1$ then $7^y+1=2^z$. If $2\mid y$ then in $\mathbb{Z}/4\mathbb{Z}$ we have $2^z=(-1)^y+1=2$, so that $y=1$, but $2=7^y+1\ge 7^1+1$ is absurd. Otherwise, $2\nmid y$, and in $\mathbb{Z}/16\mathbb{Z}$ we have $2^z=7^y+1=7\cdot 49^{\frac{y-1}{2}}+1=7+1$: this is enough to find the other trivial solution $(0,1,3)\in S$. 3. If $x\ge 1$ and $y=0$ then $5^x+1=2^z$. Since $5^x+1\ge 6$ we have that $z\ge 2$, and taking $\mathbb{Z}/4\mathbb{Z}$ we obtain $0=2^z=5^x+1=2$, absurd. 4. From now, we can suppose $\min\{x,y\}\ge 1$, so that $2^z=5^x+7^y\ge 5+7=12$ implies that $z\ge 4$. 5. In $\mathbb{Z}/3\mathbb{Z}$ we have that $(-1)^z=2^z=5^x+7^y=(-1)^x+1$, so $2\mid x$ and $2\nmid z$. 6. In $\mathbb{Z}/4\mathbb{Z}$ we have that $0=2^z=5^x+7^y=1+(-1)^y$, so $2\nmid y$. 7. In $\mathbb{Z}/16\mathbb{Z}$ we have that $0=2^z=5^x+7^y=25^{\frac{x}{2}}+7\cdot 49^{\frac{y-1}{2}}=$ $9^{\frac{x}{2}}+7=3^x+7$ but residues of $3^i$ are exactly $[3,9,11,1]$, so that we can conclude that $4\mid x-2$. 8. To sum up all observations in points 5,6,7 , we can reformulate the problem as "find all $(a,b,c)\in \mathbb{N}^3$ such that $5^{4a+2}+7^{2b+1}=2^{2c+1}$ for some $c\ge 2$. 9. In $\mathbb{Z}/25\mathbb{Z}$ we have that $0=5^{4a+2}=2^{2c+1}-7^{2b+1}=2^{2c+1}-7\cdot (-1)^b$; and since $\text{gcd}(13,25)=1$ then also $0=13(2^{2c+1}-7\cdot (-1)^b)=4^c-16\cdot (-1)^b$. Now residues of $4^i$ are exacty $[4,16,14,6,24;21,9,11,19,1]$, and we are searching $16\cdot (-1)^b \in \{9,16\}$, that is equivalent to $5\mid c-2$. 10. To sum up all observations in points 8,9 , we can reformulate the problem as "find all $(a,b,d)\in \mathbb{N}^3$ such that $5^{4a+2}+7^{2b+1}=2^{5(2d+1)}$. We divide the rest of the solution in two cases: PART 1. 11. Let's find all solutions in the case $b=0$, that is $5^{4a+2}+7=2^{10d+5}$ in non-negative integers. 12. The equation in point 11 is equivalent to $5^2\cdot (5^{4a}-1)=2^5\cdot (2^{10d}-1)$, and we can see the trivial solution with $a=d=0$ that is $(x,y,z)=(2,1,5) \in S$. Let's show that no other integer solutions exist. 13. Consider the 2-adic valuations: it is true that $5=\upsilon_2(2^5(2^{10d}-1))=\upsilon_2(5^2(5^{4a}-1))=$ $\upsilon_2(5^{4a}-1)$. And by the lifting lemma it is equal to $\upsilon_2(\frac{5^2-1}{2})+\upsilon_2(4a)$, that implies $5=2+\upsilon_2(4a)$, so $2||a$. 14. Consider the equation in $\mathbb{Z}/13\mathbb{Z}$ (after observing that $13\mid 5^4-1$), and we obtain: $6\cdot 10^d=2^5\cdot 2^{10d}=5^2\cdot 5^{4a}+7=5^2+7=6$ that is equivalent to $6=\text{ord}_{13}(10)\mid d$ 15. To sum up points 13,14, the equation becomes equivalent to $5^2\cdot (5^{2^3\cdot (2h-1)}-1)=2^5(2^{2^2\cdot 3\cdot 5k}-1)$ for some $h,k$ positive integers. 16. After observing that $17\mid \text{gcd}(5^8+1,2^4+1)$ (that is equivalent to $\text{ord}_{17}(5)=2\text{ord}_{17}(2)=16$ ), consider the equation of point 15 in $\mathbb{Z}/17\mathbb{Z}$: we have that $5^2(5^{2^3\cdot (2h-1)}-1)=5^2((-1)^{2h-1}-1)=-2\cdot 5^2=1$ and $2^5(2^{2^2\cdot 3\cdot 5k}-1)=15((-1)^k-1) \in \{0,4\}$. This is a contradiction. PART 2. 17. Now let's show that in the case $b\ge 1$, the equation $5^{4a+2}+7^{2b+1}=2^{10d+5}$ has no solutions. 18. In $\mathbb{Z}/49\mathbb{Z}$ (the information in this point makes the difference from the previous part) we have $37^a=50\cdot 37^a=2\cdot(5^2\cdot 5^{4a})=2(5^{4a+2}+7^{2b+1})$ $=2(2^{10d+5})=64\cdot 2^{10d}=15\cdot 44^d=37^{18+16d}$. Now, since $\text{ord}_{49}(37)=21$, we conclude that $21\mid (18+16d)-a$, that is equivalent to $3\mid a-d$ and $7\mid (2d+4)-a$. 19. Observe that $7^6-1=2^4\cdot 3^2\cdot 19\cdot 43$ and take the equation in $\mathbb{Z}/9\mathbb{Z}$, we have: $4^a+4^b=4\cdot (7\cdot 4^a+7\cdot 4^b)=4\cdot (25\cdot 5^{4a}+7\cdot 49^b)$ $=4\cdot (2^5\cdot 2^{10d})=2^7\cdot 2^{10d}=2\cdot 4^{2d}$. We can see directly by hand that residues of $4^i$ are exactly $[4,7,1]$ and residues of $2\cdot 4^{2i}$ are $[5,8,2]$. And we can see that $(a,b,d)$ is a solution if and only if $3\mid a+b-d$. 20. Since we know that $3\mid a-d$ for point 18, then it is also true that $3\mid (a+b-d)-(a-d)=b$. In particular if $d$ is a divisor of $7^6-1$ then $7^{2b+1}$ is constant in $\mathbb{Z}/d\mathbb{Z}$. 21. In $\mathbb{Z}/43\mathbb{Z}$ we have $25\cdot 23^a+7=5^{4a+2}+7^{2b+1}=2^{10d+5}=32\cdot 35^d$. Now $\text{ord}_{43}(23)=21$ and $\text{ord}_{43}(35)=7$: complete residues of LHS are $[23,31,0,18,2,21,28; 17,22,8,30,20,5,4; 24,11,13,16,42,38,32]$, and complete residues of RHS are $[2,27,42,8,22,39,32]$. 22. Looking previous residues we can see that if $(a,d)$ is a solution modulo $7$ ( we can say that because $7=\text{ord}_{43}(35)\mid \text{ord}_{43}(23)\mid \varphi(43)$ ) then $(a,d)\in\{(5,1),(5,3),(3,4),(2,5),(7,7)\}$. 23. In noone of previous couples $(a,d)$ modulo $7$ we verify that $7\mid (2d+4)-a$, as requested in point 18. This is a contradiction. [] Paolo Leonetti