No. $29| 2^a+3^b-2^{a-b}(2^b+5^a)-3^{a-b}(3^b+5^b)=3^{a-b}(5^a+5^b)$, hence $29|5^a+5^b$. Analogously $29| 3^a+3^b, 29|2^a+2^b.$ Then $29|(5^a+5^b)-(5^a+2^b)$, hence $5^b \equiv 2^b \pmod{29}$, and analogously we get $5^b \equiv 2^b \equiv 3^b \pmod{29}$, and $5^a \equiv 3^a \equiv 2^a \pmod{29}$. Notice that this means that $a,b$ are multiples of $\phi(29)=28$, but this leads to a contradiction since $2^a+3^b \equiv 1+1 \equiv 2 \pmod{29}$. $\blacksquare$

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Modulo 29 we have $2^b3^a(2^a+3^b)-6^a(2^b+5^a)-6^b (3^a+5^b)\equiv 0$. Hence $6^a5^a \equiv -6^b5^b$ or $30^{a-b}\equiv -1$. But $30\equiv 1$, not possible.