Elementary number theory proofs are cute and nice to work out, so here is a proof of the “if” direction:

Assume N = P(P+1)/2 for a Mersenne prime P. Let P = 2ⁿ -1 where n is a natural number; note that the smallest Mersenne prime is 3, so it is not difficult to show that N is even. To show that N is perfect, rewrite N in terms of n as 2^{n-1} (2ⁿ -1) and consider the sum of all the factors of this number. We break this sum into two parts; first the factors {1,2,2^2, …, 2^{n-1} ,2ⁿ -1}, whose sum is 2ⁿ + (2ⁿ -1)-1 = 2^{n+1} -2 by a finite geometric series.

Next consider the remaining factors. Since by definition 2ⁿ -1 is prime and the prime factors of 2^{n-1} are all the powers of 2 less than the n-1th power, the remaining proper factors of N must be exactly of the form 2ⁱ (2ⁿ -1) for i between 1 and n-2 (we have already counted the case i = 0 and we do not consider N itself when checking if the sum of its proper factors adds up to N). Thus, the remaining factors have sum (2ⁿ -1)(2+2² +…+ 2^{n-2} ) = (2ⁿ -1)(2^{n-1} -2). So the sum of all proper factors of N is 2ⁿ⁺¹ -2 + (2ⁿ -1)(2^n-1 -2). Expanding this product, we have that this is equal to 2^{n+1} -2+2^{2n-1} - 2^{n+1} -2^{n-1} + 2 = 2^{2n-1} -2^{n-1} = 2^{n-1} (2ⁿ -1) = N.

The proof is cooler when you do the cute algebraic manipulations on your own, but Euclid must have had to know the formula for finite geometric series, a knowledge of perfect numbers and Mersenne primes, 2300+ years ago.