Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$

If we write the numbers in binary system we get $$1\underbrace{0\ldots0}_a+1\underbrace{0\ldots0}_b+1\underbrace{0\ldots0}_c=1\underbrace{0\ldots0}_{2d}$$ This is possible only if $b=c$ to get $1+1=10$ and $a=b+1$ to make the $1$ dissapear with the carry.

In addition, $a$ must be odd.

Obviously, $d>0$. WLOG, assume that $a\ge b\ge c$

$2^c(2^{a-c}+2^{b-c}+1)=2^{2d}$

$2^{a-c}+2^{b-c}+1\ge 3$. Therefore, $b-c$ must be $0$ as otherwise $2^{a-c}+2^{b-c}+1$ will be an odd number larger than $1$.

$2^{a-c}+2^{b-c}+1=2^{a-c}+2=2(2^{a-c-1}+1)$ and hence $2^{a-c-1}+1$ must be even. $a-c=1$.

We have $2^c\times 4=2^{2d}$ and hence $b=c=2d-2$, $a=2d-1$.

I'd write this as $$\frac1{2^r}+\frac1{2^s}+\frac1{2^t}=1$$ where $r=2d-a$, $s=2d-b$ and $t=2d-c$ are integers. Each of $r$, $s$ and $t$ must be positive, and they can't be all $\ge2$, since then the LHS is $\le 3/4$. So one of them, say $r$, equals $1$. In that case $$\frac1{2^s}+\frac1{2^t}=\frac12.$$ The only possibility here is $s=t=2$. So the solutions for $(r,s,t)$ are $(1,2,2)$, $(2,1,2)$, $(2,2,1)$.

What about $(a,b,c)$? If $(r,s,t)=(1,2,2)$ then $(a,b,c)=(c+1,c,c)$ and $2^a+2^b+2^c=2^{c+2}$. Then $c=2d-2$ must be even, and $(a,b,c) =(2d-1,2d-2,2d-2)$ etc.