If $d>0$ then $7^a\equiv 4^b+5^c+6^d\equiv 1+5^c+0\pmod3\Leftrightarrow 5^c\equiv 0\pmod 3$,contradiction. So $d=0$ and $7^a=4^b+5^c+1$. If $b>0$ then $7^a\equiv 4^b+5^c+1\equiv0\pmod 2$,contradiction So $b=0$ and $7^a=5^c+2$. If $c\geq 2$ then $7^a\equiv 2\pmod 25$. But $ord_{25}(7)=4$ so $7^a\equiv 7,-1,-7,1\not \equiv 2 \pmod {25}$. So $c=0,1$. $c=0\Rightarrow 7^a=3$,contradiction $c=1\Rightarrow 7^a=7\Leftrightarrow a=1$. So the only solution is $(a,b,c,d)=(1,0,1,0)$

$7^a=4^b+5^c+6^d$ By Mod(4) there are two cases $\boxed{\text{ a=odd ; d=1}}$ $\boxed{\text{ a=even ; d greater than 1}}$ $\text{by mod 3 the a=odd,d=1 case is contradiction}$ $\text{by mod 5 the a=even,d greater than 1 case is contradiction}$ And we have one heavy claim if the 2 cases are false 1 case reamin this case is d=0 case $7^a=4^b+5^c+1$ By mod 4 $\implies$ $\boxed {b=0}$ $7^a=2+5^c$ By mod 25 c=1,a=1 So, $(a,b,c,d)=(1,0,1,0)$