Theorem: There are infinitely many HSSN.
1) According to the definition, an HSSN (let us also call it c^2) can be represented in the following manner c^2 = a^2 + b*(b+1)/2. The formula looks pretty much like the Pythagorean formula c^2 = a^2 + b^2. There are infinitely many Pythagorean triples (a, b, c). What if b^2 was also a triangular number?
2) There exist Square triangular numbers (STN), i.e. triangular numbers that are also square numbers (OEIS A001110), and there are infinitely many of them. It is worth noting that every STN, written as a(n), is an odd number when n is odd and an even number when n is even, as a(n) = 34*a(n-1) – a(n-2) + 2. Therefore, there are infinitely many odd and infinitely many even square triangular numbers.
3) If STN is an odd number (let us call it STNodd), then √STNodd is an odd number. Therefore, √STNodd is of the form m^2-n^2, as any odd number can be represented as the difference between two consecutive square numbers (in other words, the p-th odd number equals the p+1-th square number minus the p-th square number).
4) It follows from 3) that for each and every √STNodd there is always a unique couple of positive integers m and n (where m>n) that can be used to generate a unique Pythagorean triple with a unique hypotenuse. According to the Euclid’s formula the triple is (m^2 - n^2, 2*m*n, m^2 + n^2). Therefore, each and every of the infinitely many STNodd>1 can be paired with a unique HSSN, i.e.
c^2 = (m^2 + n^2)^2, from which we can conclude that the number of the HSSN is infinite. QED
There are many more examples of HSSN that we have not considered so far, i.e. HSSN that are sums of a positive square number and a positive triangular number that is not a STN. For example, 4=2^2=1^2+2*3/2 or 16=4^2=1^2+5*6/2.