<s>[INST] Prove that $7^x=1+y^2+z^2$ has no solutions in positive integers $$7^x=1+y^2+z^2$$
So far I have not gotten any remarkable results. Analyzing $mod$ $3$ I get that $y$ and $z$ must be divisible by $3$. Further analyzing $mod$ $4$ I got that $x$, $y$, $z$ must all have the same parity. Looking at the equation $mod$ $8$ I got that all the variables have to be even. Then looking at the equation $mod$ $9$ I got that $x$ is, in fact, divisible by $3$. So the most I know about the variables is that all of them are divisible by $6$.
Rewriting the equation as $(7^k-1)(7^k+1) = y^2+z^2$ we see that $(7^k-1)$ and $(7^k+1)$ greatest common divisor is $2$ so that means that the highest power of $3$ that divides the sum of squares must also divide $(7^k-1)$ because the other factor is not divisible by $3$.
That is how far I have gotten with this problem. 
Any ideas? 
 [/INST] $(7^{2k+1}-1)/2$ is an odd integer congruent to $3\bmod 4$ and hence has a prime of the form $4k+3$ appearing an odd number of times.
This implies there are no solutions for odd values of $x$, and we can induct over $v_2(x)$ and use the factorization $(7^{k}-1)(7^k+1)$ to get there are no solutions for any $x$ (you need to use the fact that the only prime that can divide both the numbers is $2$).

To prove the claim we just write the expression as $\frac{(7-1)(7^{2k} + \dots + 7^0)}{2}=3(7^{2k} + \dots + 7^{0}) \equiv 3\times 1 \bmod 4$
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