Topology: Polygonal Meshes, Euler's Formula
Let g be the number of vertices of valence 4. A closed sphere has G=0 and we have the equation V-E+F=2. Now V=x1+g+x2 and F=275. 

As a first step let us count the edges. If we multiply each face by its valence and sum those up, we get twice the number of edges (as each edge is shared by two faces), so:

(1)   2*E = (274*4 + x3*1)


Immediately we can see that 2 divides x3.

Similarly if we multiply each vertex with its valence we get again twice the number of edges, as each edge is shared by two vertices, so:

(2)   2*E = 3*x1 + 4*g + 5*x2


And again we see that x1 and x2 must have the same parity.

Plugging the first of this equation (with the very first ones) into V-E+F=2 we get

(3)   2 = V-E+F = x1 + g + x2 - (1096+x3)/2 + 275 = x1 + x2 - x3/2 + g + 548


Equating the two equations we derived above we get 

(4)   1096 + x3 = 3*x1 + 4*g + 5*x2


Now the system of equations (3)&amp;(4) has all the information encoded (excluding the parity information), so it appears you don't already have everything you need to know to determine these four unknowns.

I made a small Haskell program that does some bruteforcing with these conditions: Try it online!

The first solution it does get is (0,732,2564,0) = (x1,x2,x3,g)