Multiply through by 4 and factor out v-w on the left:

(v-w) * (4v^2+4vw+4w^2) = 4vw + 244

Now we'll make the substitution x=v-w and y=v+w.  Note that this also means 2v=y+x and 2w=y-x, which makes the substution easier.

x * ((y+x)^2+(y-x)*(y+x)+(y-x)^2) = (y-x)*(y+x) + 244

Now simplify:

x^3 + x^2 + 3xy^2 - y^2 = 244.

Then multiply through by 27 and subtract 4 from each side:

27x^3 + 27x^2 - 4 + 81xy^2 - 27y^2 = 6584

Now the cubic in x and mixed terms in y^2 both have 3x-1 as a factor.  So factor everything on the left side and divide by 3x-1.  Then:

(3x+2)^2 + 27y^2 = 6584/(3x-1)

 Then 3x-1 must be a factor of 6584.  Also since (3x+2)^2 + 27y^2 is strictly positive then 3x-1 must also be positive.

6584 = 2^3*823, so 3x-1 can be one of 1,2,4,8,823,1646,3292,6584.  But looking mod 3, 3x-1 must be 2 mod 3.  That leaves 2, 8, 1646 or 6584.  Then x equals 1, 3, 549, or 2195.

Testing each case:

If x=1 then y^2=121, which makes y=11 or -11

If x=3 then y^2=26: rejected as y^2 needs to be square

If x=1646 then y^2=-100711: rejected as y^2 needs to be positive

If x=6584 then y^2=-1606984: rejected as y^2 needs to be positive

So this leaves us with x=1 and y=11 or -11.

In the case x=1 and y=11 then v=6 and w=5

In the case x=1 and y=-11 then v=-5 and w=-6.

Then over all integers there are only two solutions (v,w) = (6,5) or (-5,-6).