$2^{x}+ 5^{y}= 7^{z}$ Help me solve this integers exponents equation:
$$2^{x}+ 5^{y}= 7^{z}$$
We have: $z= \frac{i(2\,\pi\,n-i\,\log(2^{x}+ 5^{y}))}{\log(7)}$
I ask because I read that a slightly complication of the equation occuring in Fermat's last theorem can lead to an undecidable case and I wonder whether the given form already is sufficient to achieve this.

Only $x=y=z=1$, corresponding to $2+5=7$, can work.  Most of this proof rehashes the comments, the real purpose here is to consolidate all the steps into one place.
1)  $x$ is odd, or else $2^x \equiv 7^z \equiv 1 \bmod 3$ and $5^y$ can't be divisible by $3$.
2)  $y$ is odd, or else $5^y \equiv 7^z \equiv 1 \bmod 3$ and $2^x$ can't be divisible by $3$.
3)  If $x$ and $y$ are both odd and $x\ge 3$, then $2^x+5^y  \equiv 0+5=5 \bmod 8$ but $7^z \in \{1,7\} \bmod 8$.  This forces $x=1$.
4)  If $x=1$ and $y\ge 2$, then $2^x+5^y  \equiv 2+0=2 \bmod 25$ but $7^z \in \{1,7,18,24\} \bmod 25$.  This forces $y=1$ leaving $2+5=7$ as the only possible equality.
