No. of solution

2013 = 3*271
271 and 3 are of the form 4k+3 while the required condition for a number n to be a sum of 2 squares is that:
n = 2^ma²b
where m≥0,a is odd, and b and every prime divisor of b is of the form 4k+3.

Since 2013 does not satisfy the conditions no solution exist.

In the previous answer I mentioned that b should have prime divisors of the form 4k+3 when it should actually be 4k+1.
and,2013 = 3*11*61.

@Swarna Kamal Dhyawala:Thanks for rectifying the mistake.
2013 = 2⁰*1²*2013

Here m=0, a =1, and b =2013 = 3*11*61

Since all the prime divisors of b are not of the form 4k+1,it cannot be written as a sum of 2 squares.

@Soumyabrata Mondal: If the number is prime it would mean that m =0,a =1 and b is a prime.
Now if b is not of the form 4k+1 it cannot be written as a sum of 2 squares.Otherwise Solutions to x and y exist.