Taken from the example on this site, I think this would be the most natural way of doing it, by filling in the header value and passing the header to the template.

This is to fill in the header Authorization:

String plainCreds = "willie:p@ssword";
byte[] plainCredsBytes = plainCreds.getBytes();
byte[] base64CredsBytes = Base64.encodeBase64(plainCredsBytes);
String base64Creds = new String(base64CredsBytes);

HttpHeaders headers = new HttpHeaders();
headers.add("Authorization", "Basic " + base64Creds);

And this is to pass the header to the REST template:

HttpEntity<String> request = new HttpEntity<String>(headers);
ResponseEntity<Account> response = restTemplate.exchange(url, HttpMethod.GET, request, Account.class);
Account account = response.getBody();

To pass basic authentication parameters in WebClient, you can use the BasicAuthenticationInterceptor class provided by Spring Security. Here's how you can modify your code to include basic authentication:

URI url = new URI("http://hostname:8080/api/route");
String username = "username";
String password = "password";

WebClient.Builder webClientBuilder = WebClient.builder()
        .defaultHeader(HttpHeaders.CONTENT_TYPE, MediaType.APPLICATION_JSON_VALUE)
        .filter(basicAuthentication(username, password));

WebClient webClient = webClientBuilder.build();

String requestBody = ""; // The body does not matter for this example

try {
    String response = webClient.post()
            .uri(url)
            .bodyValue(requestBody)
            .retrieve()
            .bodyToMono(String.class)
            .block(); // Use block() instead of toFuture().get()
    
    // Process the response
} catch (Exception e) {
    logger.error(e.toString());
}