Yes, each scored goal has probability 1/3 of being scored by team 1 and probability 2/3 of being scored by team 2, and by the magics of Poisson proceses, each goal is colored independently from the others.
Hence you look for the probability of a random walk starting from (3,1) and with step (+1,0) with probability 1/3 and (0,+1) with probability 2/3 to reach the line (5,something) before reaching the line (something,5).
Writing u(x,y) for this probability starting from (x,y), you look for u(3,1) and you know that u(5,something)=1, u(something,0)=0 and 3u(x,y)=u(x+1,y)+2u(x,y+1). This shows that u(3,1) is part of the unique solution of a linear system of equations whose unknowns are the u(x,y) for (x,y) any of the points (3,1), (4,1), (3,2), (4,2), (3,3), (4,3), (3,4), (4,4).
Starting backwards, one gets u(4,4)=1/3, u(3,4)=u(4,4)/3=1/9, u(4,3)=1/3+2u(4,4)/3=5/9, u(3,3)=u(4,3)/3+2u(3,4)/3=7/27, u(4,2)=1/3+2u(4,3)/3=19/27, u(3,2)=u(4,2)/3+2u(3,3)/3=11/27, u(4,1)=1/3+2u(4,2)/3=65/81, and finally u(3,1)=u(4,1)/3+2u(3,2)/3=131/243.
Edit An alternative solution is, as suggested by the OP, to consider the five first goals and to note that team 1 loses the game if and only if team 1 scores one or zero of these five goals. The number of goals scored by team 1 amongst the five first goals is binomial (5,1/3), hence the probability that team 1 wins is 1-P(Bin(5,1/3)=0)-P(Bin(5,1/3)=1). This is 1-(2/3)^5-5(2/3)^4(1/3)=1-32/243-80/243=131/243.