Find positive integers a, b and c that satisfy:

a!b! = a! + b! + 2^{c}

Consider first the case where a = b. Then we have:

(a!)^{2} = 2a! + 2^{c}

a! divides the left hand side, and the first term on the right hand
side, so it must divide 2^{c}. But the factorial of any number
greater than 2 includes a factor of 3, and so cannot divide a power
of 2.
Hence, a = 1 or a = 2, and correspondingly, (a!)^{2} - 2a! = -1 or 0,
neither of which is a power of 2.
So there's no solutions where a = b.

Without loss of generality, suppose that a < b, and let z = (a+1)(a+2)...(b), so that b! = za!. Now we have:

z(a!)^{2} = (z+1)a! + 2^{c}

Again, a! divides the left hand side, and the first term on the right
hand side, so a! must divide 2^{c}, and a must equal 1 or 2.
When a = 1, we get:

b! = 1 + b! + 2^{c}

which clearly has no solution. When a = 2:

2b! = 2 + b! + 2^{c}

or

b! = 2(2^{c-1} + 1)

If b > 3, then

2.3.4...b = 2(2^{c-1} + 1)

or

3.4...b = 2^{c-1} + 1

But 4 cannot divide 2^{c-1} + 1, so this case is impossible.
Hence b must equal 3, since it's less than 4 and greater than a, which
equals 2.
We get the solution a = 2, b = 3 and c = 2:

2!3! = 2! + 3! + 2^{2}