If y=2 then the equation reduces to 9=2^(x+1)+7.  This would imply x=0, but that is not a positive integer.

If y=3 then the equation reduces to 64=2*3^x+10, which has a solution at x=3.  (x,y)=(3,3) is a solution.

Now to show that there are no "large: solutions, y>=4.

Let y+1=w, for a bit of clarity.  So the "large" solutions are w>=5.  

Then the equation becomes w^(w-1) = 2*(w-1)^x + 3w-2.

Take this mod w.  Then 0 = 2*(-1)^x - 2 mod w.  If x is even then this is an identity.  If x is odd then further reduction results in 0=4 mod w.

In the x is odd case I conclude w is a factor of 4. But these are all less than 5 so no new solutions here.

For the x is even case, go back and take the equation mod w^2.

Then 0 = 2*(1-wx) + 3w-2 mod w^2.  This simplifies to 0 = (2x-3)*w mod w^2.  Since w is a factor on both sides the congruence can be stepped down to 0 = 2x-3 mod w.

From here I can handwave it.  For large w I can approximate the equation as w^(w-1) ~= 2*w^x.  From this I expect x to be roughly the same size as x.  But 2x-3 mod w implies that x is not that close, so I conclude there are no large solutions.