[nnqxrfnw] Pierpont prime powers

Consider the Diophantine equation 2^x * 3^y + 1 = p^z with p restricted to prime numbers and z>1.  Are these the only solutions?

2^0 * 3^1 + 1 = 2^2
2^3 * 3^0 + 1 = 3^2
2^3 * 3^1 + 1 = 5^2
2^4 * 3^1 + 1 = 7^2
2^5 * 3^2 + 1 = 17^2

We searched all Pierpont numbers (numbers of the form 2^x*3^y+1) less than 2^10000.

There are many solutions, namely the Pierpont primes, for z=1.

limt=1000; oldlim=0; for(i=0,limt,for(j=0,limt,x=2^i*3^j+1;if(x>2^limt,break);if(x>oldlim,if(ispseudoprime(x),print(i," ",j," 1"),a=isprimepower(x);if(a>1,print(i," ",j," ",a))))))

We avoid calling isprimepower on primes because it is slow for large primes.