So we have $7^a-4^b-5^c= 6^d,$ taking $\mod 3,$ we get for $d\ge 1,$ LHS as $1/2 \mod 3$ but RHS is $0 \mod 3 \implies d=0.$ So we have to solve $7^a-4^b-5^c=1.$ Now taking $\mod 2,$ we get $b=0 \implies 7^a - 1-5^c=1 \implies 7^a -5^c =2.$ Now for $c\ge 2,$ we have $7^a \equiv 2 \mod 25$ but $7^a \equiv 7,24,18, 1 \mod 25.$ Hence $a=c=1$ is the only solution for $7^a -5^c =2.$ So the solution is $( 1,0,1,0).$