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 for (int i = 2; i * i <= max; i++) {

In my testing it's slightly faster to take the square root of max rather than multiplying i * i on each iteration.

 for (int i = 2, limit = 1 + (int)Math.sqrt(max); i <= limit; i++) {

It's only a fraction of a millisecond but a noticeable difference.

 for (int j = i; i * j <= max; j++) {

 isPrimeArray[i * j] = false;
 }

You could simplify this.

 for (int j = i * i; j <= max; j += i) {
 isPrime[j] = false;
 }

Then rather than doing two multiplications per iteration, you can do one addition. Of course a good optimizer would notice that both multiplications are the same and only do one.

I prefer names that do not include the type, so I'd just call it isPrime.

 if(isPrime) {

 nthPrime++;
 }
 if(nthPrime == n) {
 return index;
 }

You compare nthPrime and n more often than necessary. You can just say

 if (isPrime) {
 nthPrime++;
 if (nthPrime == n) {
 return index;
 }
 }

As the result can only change when nthPrime does (or n but n never changes).

Incidentally, I tried expanding the for loop as @h.j.k. suggested but it ran more slowly than the original code. Only a fraction of a millisecond but slower.

 for (int i = 2; i * i <= max; i++) {

In my testing it's slightly faster to take the square root of max rather than multiplying i * i on each iteration.

 for (int i = 2, limit = 1 + (int)Math.sqrt(max); i <= limit; i++) {

It's only a fraction of a millisecond but a noticeable difference.

 for (int j = i; i * j <= max; j++) {

 isPrimeArray[i * j] = false;
 }

You could simplify this.

 for (int j = i * i; j <= max; j += i) {
 isPrime[j] = false;
 }

Then rather than doing two multiplications per iteration, you can do one addition. Of course a good optimizer would notice that both multiplications are the same and only do one.

I prefer names that do not include the type, so I'd just call it isPrime.

 if(isPrime) {

 nthPrime++;
 }
 if(nthPrime == n) {
 return index;
 }

You compare nthPrime and n more often than necessary. You can just say

 if (isPrime) {
 nthPrime++;
 if (nthPrime == n) {
 return index;
 }
 }

As the result can only change when nthPrime does (or n but n never changes).

Incidentally, I tried expanding the for loop as @h.j.k. suggested but it ran more slowly than the original code. Only a fraction of a millisecond but slower.